Here is a common informal proof for the chain rule:
If $S(a)=f(g(x))|_{x=a}$, then $S'(a)$ is given by
\begin{align} \lim_{x \to a}\frac{S(x)-S(a)}{x-a}&=\lim_{x\to a}\frac{f(g(x))-f(g(a))}{x-a} \\ &=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\cdot\frac{g(x)-g(a)}{x-a} \\ &=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\cdot\lim_{x \to a}\frac{g(x)-g(a)}{x-a} \\ &=f'(g(a))g'(a) \end{align}
The reason this proof is considered informal is because it does not take into account the case where $g(x)=g(a)$ as $x \to a$. This is not, however, where my confusion lies. I am confused as to how $f'(g(a))$ equals $$\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}$$
As far as I understand, $f'(g(a))$ means apply the function $f'$ to $g(a)$. The function $f'$, evaluated at a particular point $p$, is defined by
$$ \lim_{x \to p}\frac{f(x)-f(p)}{x-p} $$
Thus, $f'(g(a))$ is defined as
$$ \lim_{x \to g(a)}\frac{f(x)-f(g(a))}{x-g(a)} $$
And so I don't see how these two expressions can be equal to each other. What am I missing?
Since $g$ is differentiable in $a$ and $f$ is differentiable in $g(a)$, then they are also continuous in those points; therefore $g(x) \to g(a)$ as $x\to a$ and you can use substitution $y=g(x)$ in the limit $\lim_{x\to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)}$ to get:
$$\lim_{x\to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} \stackrel{y=g(x)}{=} \lim_{y\to g(a)} \frac{f(y) - f(g(a))}{y - g(a)} = f^\prime (g(a))\; .$$