$Var(aX) = a^2Var(X)$ but $Var(X+Y) = Var(X) + Var(Y)$. How does this make any sense? Apologies for the less than clear question, I wasn't quite sure how to phrase it. 

Say you have a random variable $X ~ N(5, 10^2)$. Say you have another random variable $Y ~ N(5, 10^2)$. 

Var(2X) = 4Var(X) = 4*10^2

But:

$Var(X+Y) = Var(X) + Var(Y) = 2\times 10^2$

But now a question arises: what?! X = Y so Var(2X) = Var(X+X) = Var(X+Y) and yet two different answers are reached using the two distinct accepted formulas. How does this make any sense? What's going wrong? It shouldn't matter what we call the random variables, and yet it seems as if it does.
Because in general
$$
\mathbb{V}{\rm ar}(X + Y) = \mathbb{V}{\rm ar}(X) + \mathbb{V}{\rm ar}(Y) + \color{red}{2\operatorname{\mathbb{C}{\rm ov}}(X,Y)}
$$
in your last step you missed to include the covariance of $X$ with $X$
X is distributed identically and independently to $Y$, but $X$ IS NOT $Y$. Let's say $X$ and $Y$ each represent a die roll. If you roll $X$ and $Y$, you might get a $2$ and a $3$, which clearly are not equal. If you compare $X$ to $X$, both are $2$. For any pair of die rolls, the value of $X$ will always be equal to the value of $X$, but will not necessarily be equal to the value of $Y$, and the same goes for any pair of iid $X$ and $Y$.