What is the difference between Hom and Sheaf Hom?

I'm reading Hartshorne's book, and in 3.6 he begins to go into detail about Ext and sheaf Ext, which are derived functors of Hom and sheaf Hom respectively. Let $\mathcal{F,G}$ be sheaves of $\mathcal{O}_X$ modules on a scheme $X.$ $Hom(\mathcal{F,G})$ is the set of $\mathcal{O}_X$ module homomorphisms between $\mathcal{F}$ and $\mathcal{G}$. Hence, if $\varphi \in Hom(\mathcal{F,G})$ then for every open set $U \subset X,$ we have a map $\varphi|_{U}$. On the other hand, the sheaf $\mathcal{Hom(F,G)}$ assigns to each open set $U\subset X$ the set of $\mathcal{O}_U$ module homomorphisms $Hom \mathcal{(F}(U),\mathcal{G}(U))$ (where $\mathcal{O}_U$ is considered as a ring, not a ringed space). It seems to me that both Hom and sheaf Hom encode the same information in different ways, so I don't understand why their derived functors seem different.

EDIT: Actually I think Hom and sheaf Hom might be different in the following way: for Hom, we are looking at "global maps," i.e. maps of $\mathcal{O}_X$ modules, which we can then restrict to an open set. On the other hand, sheaf Hom assigns to each open set a hom-set of maps, and some of those maps may not arise from restriction of a "global map." But since sheaf Hom is a sheaf, not a presheaf, this probably shouldn't be a problem, and all maps are indeed restrictions of global maps since we can glue them together. Am I right about this? -Hom is just the global sections of sheaf Hom. Thanks Qiaochu

• It sounds like you're claiming that every section of a sheaf is a restriction of a global section. This is certainly false. – Qiaochu Yuan Feb 1 '15 at 1:26
• Yea I was wrong about that. I edited my question after I saw your answer, but I didn't take away the false statement (just so I can remember that it's false). – 010110111 Feb 2 '15 at 1:40