Let me preface this question by saying I'm actually a physicist, not a mathematician, so a lot of the language I see you guys using here is over my head, so if you can keep it simple, that would be preferable.
The fundamental lemma of calculus of variations states that, if $f\in C^k[a,b]$, $\int_a^b f(x) h(x) dx = 0$, and $h(a) = h(b) = 0$, then $\forall h \in C^k[a,b]$: $f(x) = 0$ $\forall x \in [a,b]$.
The proof of this usually goes something along the lines of assuming some specific form of $h(x)$, something like $h(x) = r(x)f(x)$ where $r(x) > 0$ on $(a,b)$ and $r(a) = r(b) = 0$, substituting this into the integral, and then proving that this for the integral to remain zero, this requires that $f(x) = 0$, and proof completed. My confusion is this: does this not only mean that the theorem holds for some small subset of $h$'s, namely, those fitting the form assumed to replicate the proof, rather than the general class of $h$'s proposed in the lemma above? The proof assumes that if you can demonstrate $f(x) = 0$ for some small subset of $h$'s, then it works for a larger class of $h$'s, which I have a hard time seeing as justified. What am I missing here?
You're quoting the lemma wrong. It should be something like
In other words the $\forall h$ is in the assumptions of the lemma, not the conclusion.
The fact that only a certain few $h$s are used in the proof just means that the lemma assumes more than it strictly needs to -- which is to say that it promises less than it can keep. That can't make it less true than it would be if it listed precisely those $h$ that it needed the premise to hold for.