which is the meaning of scatterplot between a pair of 2 consecutive pseudo random numbers with respect to the independence of the sequence? Pseudo random number generators should give as output random sequences u1, u2, ... that are mutually independent and identically distribuited (iid). 
Since testing for independence is not easy, the first check is testing for noncorrelation. As a first visual test you could check the scatterplot of (Ui, Ui-1). 
The output should fill the unit square almost evenly: in this case we can conclude that Ui is incorrelated to Ui-1.
Can we also conclude that Ui is independent to Ui-1?
Why if all the sequence should be mutually independent, then Ui vs Ui-1 should span the unit square almost evenly?
If also the Autocorrelation Function signals no autocorrelation at any lag, what can we conclude? (That all the pairs Ui, Uj with i different from j are not correlated?)
Since the elements of the sequence U1, U2,... should be mutually independent in order to be iid, should we check also the correlation between all the combinations(3-tuple, 4-tuple, ... N-tuple) and not only two consecutive pair (Ui, Ui-1)? 
 A: You ask four questions.

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As a first visual test you could check the scatterplot of $(U_i, U_{i-1}).$ The output should fill the unit square almost evenly: in this case we can conclude that $U_i$ is uncorrelated with $U_{i-1}.$  Can we also conclude that $U_i$ is independent of $U_{i-1}$?

No.  First, modern pseudorandom number generators (such as those used in most statistical applications) will behave exactly as described, but because they provide deterministic sequences of values (albeit extremely long ones), they cannot be truly independent.
This argument will not convince some people, who will (rightly) argue that the lack of independence is so slight that it couldn't possibly matter.  Allow me, then, to exhibit an example involving a sequence of just three uniform random variables $(U_1,U_2,U_3)$ that behave as described but are demonstrably not independent.  Here is a scatterplot matrix of the first thousand realizations showing how the pairs $(U_i,U_{i-1})$ are uniformly filling the unit square:

However, the three variables are not independent, as this map of $U_3$ versus $U_1$ and $U_2$ demonstrates:

Here is how the variables were generated.  We begin with a set $\Omega$ of integer vectors,
$$\Omega = \{(0,0,0),\ (0,1,1),\ (1,0,1),\ (1,1,0)\},$$
and give it the uniform probability distribution (so each element is chosen with $1/4$ probability).
To create one realization of $(U_1,U_2,U_3),$ take an infinite sequence $(\omega_i),i=1,2,\ldots,n,\ldots$ of independent draws from $\Omega.$  Writing $\omega_{ij}$ for component $j$ of $\omega_i,$ set
$$U_i = \sum_{j=1}^\infty \omega_{ij}2^{-j}.$$
In effect, for each $j$ the sequence $(\omega_{ij})$ is a random string of zeros and ones which is interpreted as the binary representation of a number between $0$ and $1.$  It is obvious--and straightforward to prove--that each $U_i$ has a uniform distribution. (See Method #5 in my post at https://stats.stackexchange.com/a/117711/919 for more explanation and a simulation.)
Note, however, that the elements of $\Omega$ enjoy an unusual property: any two components of $\omega\in\Omega$ determine the third.  (The third equals $1$ when the other two are not equal and otherwise the third equals $0.$)  Thus, because almost all possible $U_i$ uniquely determine the sequence of $\omega_{ij}$ in their binary representations, with probability $1$ each is a function of the other two.  Consequently, the distribution of (say) $U_3$ conditional on $U_1$ and $U_2$ is a constant, rather than being uniform.  This is as far from independence as one can possibly get!
See the function predict3 in the appendix (below) for how the third of the $U_i$ is computed from the other two: you just represent the two values in binary, work out the corresponding binary representation for the third one, and convert that to a number.


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Why if all the sequences should be mutually independent, then $U_i$ vs $U_{i-1}$ should span the unit square almost evenly?

Independence means the joint distribution function of $(U_i,U_{i-1})$ is the product of the marginal distribution functions.  Having a uniform distribution means the chance that $U_i$ lies in an interval $[a,b]\subset[0,1]$ is $b-a.$  Thus, the chance that $(U_i,U_{i-1})$ lies within a rectangle $[a,b]\times[c,d]\subset [0,1]^2$ equals $(b-a)(d-c),$ which is the area of that rectangle.  Thus, for rectangles at least, the chances are equal to the areas: they are uniform.  A limiting argument is needed to show the distribution is truly uniform in the sense that the chance $(U_i,U_{i-1})$ lies in any arbitrary set $A\subset[0,1]^2$ of area $a$ is precisely $a.$  See https://stats.stackexchange.com/a/256580/919 for an example of how such arguments go.


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If also the autocorrelation function signals no autocorrelation at any lag, what can we conclude? (That all the pairs $U_i, U_j$ with $i$ different from $j$ are not correlated?)

Yes. That's because the sequence $U_1,U_2,\ldots,U_n,\ldots$ is stationary: the distributions of $(U_i,U_j)$ and $(U_{i+s},U_{j+s})$ are the same for any positive integer $s.$


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Since the elements of the sequence $U_1, U_2, \ldots$ should be mutually independent in order to be iid, should we check also the correlation between all the combinations (3-tuple, 4-tuple, ... N-tuple) and not only two consecutive pairs?

Yes. But even that's not enough: a generalization of the construction in the answer to question $(1)$ (changing from $3$ to $N+1$ components) provides an example of what can go wrong.   But as a practical matter, such checks are an excellent idea: they are the basis for most procedures to check random number generators.

Appendix
This R code illustrates the calculations and produces the figures.
#
# Draw a sequence of `n` vectors from Omega.
#
rb3 <- function(n) {
  z <- matrix(c(1,1,0, 0,1,1, 1,0,1, 0,0,0), 3, 4, dimnames=list(c("x1", "x2", "x3")))
  z[, sample.int(4, n, replace=TRUE), drop=FALSE]
}
#
# Generate (U[1], U[2], U[3]) up to double precision.
#
ru <- function(nbits=52) {
  rb3(nbits) %*% (1/2)^(1:nbits)
}
#
# From two components (x,y) of (U[1], U[2], U[3]), predict the third.
#
predict3 <- function(x,y, nbits=52) {
  #--Convert a float between 0 and 1 into its binary representation
  to.binary <- function(z) {
    a <- integer(nbits)
    for (i in 1:nbits) {
      z <- 2*z
      a[i] <- floor(z)
      z <- z - a[i]
    }
    a
  }
  #--Convert a binary representation into a float between 0 and 1.
  from.binary <- function(a) sum(a * (1/2)^(1:nbits))
  
  from.binary(to.binary(x) != to.binary(y))
}
#
# Conduct a simulation of (U[1], U[2], U[3])
#
set.seed(17)
U <- t(replicate(1e3, ru())[,1,])
# sum((U[,3] - mapply(predict3, U[,1], U[,2]))^2) # Compares U[,3] to its predictions

#-- Scatterplot matrix
pairs(U, col="#00000040", labels=paste0("U[", 1:3, "]"))
#
# The plot of U[3] vs. (U[1], U[2]).
#
library(ggplot2)
b <- 8 # Number of bits in the values
x <- seq(0, 1, length.out=2^b+1)
x <- x[-length(x)]
X <- expand.grid(U1=x, U2=x)
# Compute U[3].
# X$U3 <- apply(as.matrix(X), 1, function(u) predict3(u[1], u[2], b+1)) # Long...
# -- Alternative (instantaneous):
library(bitops)
X$U3 <- with(X, bitXor(2^b*U1, 2^b*U2)) / 2^b

names(X) <- paste0("U", 1:3)
ggplot(X, aes(U1, U2)) + 
  geom_raster(aes(fill=U3)) + 
  scale_fill_gradientn(colors=rainbow(13)[1:10]) + 
  xlab(expression(U[1])) + ylab(expression(U[2])) + 
  guides(fill=guide_colorbar(expression(U[3]))) + 
  coord_fixed() +
  ggtitle(expression(paste(U[3], " depends on ", U[1], " and ", U[2])))

A: I want to give some conceptual clarifications: 

Pseudo random number generators should give as output random sequences u1, u2, ... that are mutually independent and identically distribuited (iid). 

Pseudo random number generators give you an output, that is actually completely deterministic (which is somehow the opposite of what you write). But this deterministic sequence should looks like being random in some implicitely or explicitely defined manner.
This implies that it does not make sense to check, if the pseudo random numbers are iid samples of a specific distrubition, because you already know, they are not. 
But what you can do, is to define some criteria of behaviour of the pseudo random sequence, that if fulfilled, lets you conclude that "this pseudo random sequence looks like being drawn from an iid sample from these defined point of views".
Your proposal regarding the construction of pairs $(U_n, U_{n-1})$ is an exploration, if two consecutive numbers are correlated. This means, you particularly look at the correlation as a quality check for your pseudo random number generator (because you want the resulting sequence to looks like independent, and independence implies uncorrelatedness, which implies you see no trend if you plot above pairs). 
You could find other aspects of random appearance also interesting or even more interesting. A good pseudo random number generator looks at many different aspects. A famous battery of quality checks for a pseudo random number generator are the diehard tests: https://en.wikipedia.org/wiki/Diehard_tests 
A: You're right that this is a very difficult problem--it's hard to test independence in general without a lot of data. There are loads of test statistics you can construct and test to try to reject the null hypothesis of i.i.d. samples, but unless you guess the mechanism of the dependence correctly or have a huge sample size, such statistics (like you describe in your problem) can have very low power.
If $U_1,U_2,\ldots,U_n$ are "random" integers, you could do a chi-square test of independence on sequences of a given length $k$. Then you've got $n-k$ observed sequences, and each should occur with probability $10^{-k}$.
Suppose you have some a priori knowledge of the dependence; for instance, you suspect that an adversary is trying to make the sequence look independent to the human eye, and so they make sure every number in the sequence is different from the last. Then using $k=2$ in the previous paragraph will uncover the dependence, because the 10 "doubles" will never appear in the sequence and the chi-square test for independence will reject. But if the dependence mechanism is something else, this test may not pick it up.
A: correlation will get you nowhere since correlation 0 can be yield for 2 variables that will form V-shape in scatter plot, so it's bad idea to say $corr==0 <=> 2$ variables are independent
all you can do is to produce tupples (dimension N) and ten do a monte-carlo: generate dozens of samples and then check is vectors are distributed evenly in your (N dimensional) sample space
