One simple way to think about probabilities is to enumerate all the possibilities and weight them according to their likelihoods, and then you can just count the number of different combinations. (This approach will work when the events are discrete and few in number.)
Let's start with a fair 4-sided die (you'll see the reason for this odd choice soon) and a fair coin. By "fair", I mean that all the possible outcomes are equally likely. That is, the likelihood of getting a 1 is exactly the same as getting a 4, for example, and that the likelihood of getting heads is exactly the same as getting tails. As @MichaelChernick points out, if you were to flip the coin 4 times and count the number of heads, you would find there are 5 possible final counts, so we will equalize the ranges by flipping the coin only 3 times, and subtracting 1 from the result of the die roll. Thus, both will range from 0 to 3. Now we can lay out the possibilities.
For the die:
{0, 1, 2, 3}
All of which are equally likely, because we've stipulated the die is fair, and so each possible event has a probability of 1/4.
For the coin, it's more complicated. Each individual flip is {H, T}, but we ultimately want to know about the combination of 3 different flips, so we need to lay out the combinatorics and then count the resulting number of heads.
flip
possibility 1 2 3 number of heads
1: {H, H, H, 3
2: H, H, T, 2
3: H, T, H, 2
4: H, T, T, 1
5: T, H, H, 2
6: T, H, T, 1
7: T, T, H, 1
8: T, T, T} 0
We see here that the number of possibilities grows exponentially with each additional flip. Specifically, it doubles, because there are two possible outcomes of each flip; if we were flipping a 3-sided coin (assuming one could exist) it would triple. If we had thrown a 6-sided die and flipped the coin 5 times, it would have taken 32 lines to lay out all the possible combinations of outcomes. In our case, with three flips, it took $2^3=8$ lines to lay out all of the possibilities.
Now (again) because we stipulated that heads and tails are equally likely, and because we have laid out all the combinations, each possibility is equally likely, having a probability of 1/8. But note that this does not mean that each possible number of heads is equally likely. There are three different ways to end up getting 1 head, and three ways to end up with 2 heads, but only 1 way each of getting 0 or 3 heads. Thus, the probability of getting 0 heads is 1/8, of getting 1 head is 3/8, of getting 2 heads is 3/8, and of getting 3 heads is 1/8. In general, the probability of getting the number of "successes" $k$ (i.e., heads in our case) out of the number of "trials" $n$ (flips), where each trial has a probability of success $p$, will be:
$$
p(k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}
$$
As a result of this, "the coin tosses sort of lump in the middle", as you suspect, and as you can clearly see:
These distributions do have things in common (which I think is what is confusing you): they are both discrete, have the same number of possible results (at least under our modified arrangement), and have the same average (more technically expected value). However, as their Wikipedia pages state, they differ in their higher moments; namely their variances (uniform: $(n^2-1)/12$ vs. binomial: $np(1-p)$ ) and kurtoses (uniform: $-(6n^2+6)/(5n^2-5)$ vs. binomial: $(1-6p+6p^2)/(np(1-p))$ ) differ.
On a slightly different note, the title of the question asks about how the distribution of coin flips will compare to the normal distribution, while the body of the question asks about how it compares to a discrete uniform. My answer here has mostly focused on the body of the question, but I should mention that the binomial is sometimes colloquially referred to as the "discrete normal" distribution. One way to think about this is that as the number of coin flips continues on towards infinity, it's binomial distribution will converge to the (continuous) normal distribution.
