Derivation of the variance of the sampling distribution of the mean - one unclear sentence in an explanation From OnlineStatBook: 

I don't understand the meaning of 

Since the mean is $\frac{1}{N}$ times the sum, the variance of the sampling distribution of the mean would be $\frac{1}{N^2}$ times the variance of the sum, which equals $\frac{σ^2}{N}$.

I only recently started refreshing my knowledge of statistics, and this sentence stumps me. I understand that the mean of any set of measurements is 1/N, but why does it appear squared here?  
 A: I think this can summarized by noting that $\text{Var}(cX)=c^2\text{Var}(X)$. 
The $1/N$ in the mean, ends up outside the variance parenthesis as $1/N^2$.
So
$$\text{Var}\left[\frac{1}{N}\sum_{i=1}^N X_i\right]=\frac{1}{N^2}\text{Var}\left[\sum_{i=1}^N X_i\right]\underset{\small\begin{matrix}by\\variance\\sum \,law\end{matrix}}{=}\frac{1}{N^2}N\sigma^2=\frac{1}{N}\sigma^2$$

Following up on the first comment:
\begin{align}\text{Var}[cX]&=\mathbb E\left[\left(cX-\mathbb E[cX]\right)^2\right]\\&=\mathbb E\left[\left(cX-c \,\mathbb E(X)\right)^2\right]\\&=c^2\, \mathbb E\left[(X-\mathbb E(X))^2\right]\\&=c^2\text{Var}(X).\end{align}
A: The sample $X_1,\dots,X_n$ of size $n$ is assumed to be iid (idependent and identically distributed) with mean $\mu$ and variance $\sigma^2$, i.e. $E(X_i) = \mu$ for every $i$ and $V(X_i) = \sigma^2$.
Consider the sample mean $$\bar X = \frac{1}{n}\sum_{i=1}^n X_i.$$
Expected value of that term is given by $$E(\bar X) = E\left(\frac{1}{n}\sum_{i=1}^nX_i\right) = \frac{1}{n}\sum_{i=1}^n E(X_i) = \frac{n}{n}\mu = \mu$$
where it was used that $E$ is linear (hence constant and sum can be taken out of the expectation) and that every $X_i$ has the same mean $\mu$. 
The variance of the sample mean can be calculated similarily:
$$V(\bar X) = V\left(\frac{1}{n}\sum_{i=1}^nX_i\right) = \frac{1}{n^2}\sum_{i=1}^nV(X_i) = \frac{n}{n^2}\sigma^2 = \frac{\sigma^2}{n}.$$
The variance acts a little different than expecation though. The variance is not linear in its argument but homogen of degree 2, i.e. $V(aX) = a^2V(X)$. Furthermore, the variance of a sum of random variable is only the sum of variances if the random variables are pairwise uncorrelated (i.e. in particular if the random variables are independent). 
I recommend going through the steps I provided and ask if there remains something unclear.
By the way, the sentence you don't understand the meaning of is not quite correct either as it asserts that the variance of a sum of random variables is the sum of random variables (times $1/n^2$). But as I stated this is only correct in the case of pairwise uncorrelated, or even stronger, idependent random variables.
