Scaling the normal distribution? I might just be slow (or too drunk), but I'm seeing a conflict in the equations for adding  two normals and scaling a normal. According to page 2 of this, if $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$, then $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$, and for some $c \in \mathbb{R}$, $cX_1 = N(c\mu_1, c^2\sigma_1^2)$.

Then for $X \sim N(\mu,\sigma^2)$, we have $X + X = N(\mu + \mu,\sigma^2 + \sigma^2) = N(2\mu,2\sigma^2)$, but also $X + X = 2X = N(2\mu,2^2\sigma^2) = N(2\mu,4\sigma^2)$ ? Ie, the variances disagree.

edit: Oh, am I mistaken in saying that $2X = X + X$? Is the former "rolling" $X$ just once and doubling it while the latter "rolls" twice and adds them?
On the first page of the cited document, $X_1$ and $X_2$ were previously defined to be two (distinct) independent, identically distributed random variables. For your purposes, the "identically distributed" part is not important, but the "independent" part is.

On the second page, where $X_1$ and $X_2$ are considered to be normal variables, there's still the assumption that they're independent. Possibly this could have been stated more clearly, but in context this assumption makes sense.

When you consider $2X = X + X$, you are not dealing with two independent variables.
The two "copies" of $X$ are correlated (in fact, as correlated as any two variables can be).
The formula for the sum of two independent normal variables therefore does not apply.
You make a mistake assuming that $X+X = 2X$ like in arithmetic. This does not hold for random variables. As a simple counterexample take $X$ to be a Bernoulli trial with succes probability $0.5$. If $X+X=2X$ then you cannot flip two coins and get heads on one and tails on the other. Summing random variables is given by a convolution.