For what it's worth, here is a redacted version of my inner monologue upon being exposed to your first example
$$\mathrm{Hom}_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F})\cong \mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{G},f_*\mathcal{F})$$
"So, it's two hom-sets that are basically equal, which ones? Oh wait, first note they are different kinds of homs, one is sort of over $X$ and the other over $Y$; ah I see, there is probably a map -- yup, that's the $f$ inside there -- and probably something gets pushed forward and/or pulled back with that map. Ah, I see, it's those sheaf things, never liked them, but basically I guess it's like modules, so let's think of $\mathcal F$ and $\mathcal G$ as modules, and there's a map $f$ between, uh, either those modules or their rings or something. Ah, I guess $\mathcal{G}$ is a $Y$-module and $\mathcal{F}$ is an $X$-module, and the $f^*$ turns $Y$-into $X$-modules and the $f_*$ the other way. And somehow that is probably something extremely natural when one writes it down, that some homomorphisms of $X$-modules can naturally be translated into homs of $Y$-modules, and back. Or at least one direction in these things is always trivial, maybe the other needs work, but it's probably not worth writing it down. So some hom-set gets identified with another hom-set, over a different ring, by abstract nonsense, probably helpful changing from one ring to another and back."
And as for
$$\mathrm{id} \times \psi^{\otimes n}:\mathcal{F}\otimes \mathcal{L}^n \vert_U \cong \mathcal{F}\vert_U$$
"What the ... ugh, that's shit notation out of context, the whole first thing has to be read together, one should put parentheses around it, like this: $(\mathrm{id} \times \psi^{\otimes n})$. And that is a map, yes? From, hmm, that tensor product, whatever that is, but obviously $id$ acts on the first factor, and on the second, that kind of $n$-fold product of $\psi$, which makes sense as $\mathcal{L}^n$ must be some $n$-th power, although now I don't understand that tensor power in the map. Whatever. Oh and it says it makes something isomorphic to just the $\mathcal{F}$, on $U$, whatever that all is. So it seems like that tensor power of that $\psi$ sort of kind of kills the second factor? Alright. Would need to look closer into this."