As said by Martin, first you have to find a morphism between the two sheaves, then you can use that solution locally. So, here is the morphism.
Consider on $Y$ the presheaf $P$ with sections $V\mapsto \mathscr{F}(f^{-1}(V))\otimes\mathcal{E}(V)$ for all $V\subseteq Y$ open. The sheafification of $P$ is $f_*(\mathscr{F})\otimes\mathcal{E}$.
Similarly, consider $P'$ the presheaf on $X$ with sections $U\mapsto\mathscr{F}(U)\otimes f^*\mathcal{E}(U)$. The sheafification of $P'$ is $\mathscr{F}\otimes f^*\mathcal{E}$.
Now, for all open $V\subseteq Y$, we have a moprhism $\mathcal{E}(V)\to f^*\mathcal{E}(f^{-1}(V))$, this gives a morphism
$$P(V)=\mathscr{F}(f^{-1}(V))\otimes\mathcal{E}(V)\to\mathscr{F}(f^{-1}(V))\otimes f^*\mathcal{E}(f^{-1}(V))=f_*P'(V)$$
Hence I have a natural morphism $\phi:P\to f_*P'$. Now, I have the sheafication morphism $P'\to P'^{sh}=\mathscr{F}\otimes f^*\mathcal{E}$, hence a morphism $f_*P'\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$ that, composed with $\phi$, gives a natural morphism $P\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$. Finally, passing to the sheafification of $P$, I get a morphism
$$\psi:f_*(\mathscr{F})\otimes\mathcal{E}\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$$
What a mess! Fortunately, when we restrict to an open set where $\mathcal{E}$ is free, everything looks nicer.
So, restrict to an open set $W\subseteq Y$ where $\mathcal{E}$ is free. Now, $\mathcal{E}$ and $f^*\mathcal{E}$ are free, hence you can easily check that $P$ and $P'$ are already sheaves, so $\psi=\phi=\operatorname{id}\otimes\gamma$ where we have $\gamma:\mathcal{E}\to f_*f^*\mathcal{E}$. But, for $\mathcal{E}$ free, you can easily check that this is an isomorphism.
p.s: the passage in the proof you have found not clear maybe it's simpler if viewed in this way: $f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n})\;\cong\; f_*(\mathscr{F}^{n})\cong f_*(\mathscr{F})^n$, and you can "take the $n$ out" just applying the definition of $f_*$.
$\def\H{{\mathcal Hom}}\def\HH{{\operatorname{Hom}}}$Everything you wrote seems correct for the locally free case. Though its easy to overlook subtle things with these type of arguments, it looks good to me.
For the second I believe there is a natural map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$. I will actually define a map $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$ and use the adjunction with $f^*$ to get the required map.
Let $U \subseteq Y$ be open. Then $\H_Y(E,F)(U) = \HH(E|_U,F|_U)$ and $f_*\H_X(f^*E,F^*F)(U) = \HH(f^*E|_{f^{-1}(U)}, f^*F|_{f^{-1}(U)})$. Then we note that
$$
f^*E|_{f^{-1}(U)} = f|_{f^{-1}(U)}^*\left(E|_{f^{-1}(U)}\right)
$$
and $f^*|_{f^{-1}(U)}$ is a functor from sheaves of $\mathcal{O}_U$ modules to $\mathcal{O}_{f^{-1}(U)}$ modules. Thus we get a natural map of sheaves of modules from $U$ from functoriality:
$$
\HH(E|_U, F|_U) \to \HH\left(f|_{f^{-1}(U)}^*\left(E|_{f^{-1}(U)}\right), f|_{f^{-1}(U)}^*\left(F|_{f^{-1}(U)}\right)\right) = \HH(f^*E|_{f^{-1}(U)}, f^*F|_{f^{-1}(U)}).
$$
Everything is suitably natural enough that it should commute with the restriction maps for inclusions $V \subset U$ giving a map of sheaves $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$.
Now that we have the required map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$, I think we need $X$ and $Y$ to be schemes not just locally ringed spaces. Then we can reduce to checking that this is an isomorphism on affine covers in which case it reduces to the isomorphism from commutative algebra. Without having $X$ and $Y$ be schemes I don't think we can make the argument work because $\H$ does not commute with taking stalks so we can't just check it on local rings.
EDIT: See comments below, apparently the argument does work on any ringed space as long as $E$ is of finite presentation.
Best Answer
$\def\HH{{\mathcal{H}om}}\def\Hom{{\operatorname{Hom}}}\def\N{{\mathcal N}}\def\P{{\mathcal P}}$This comes from checking the natural isomorphism from adjunction on each open set and using naturality to see it induces a morphism of sheaves not just sets.
Explicitly, take an open set $U \subset Y$. Then
$$ f_*\HH_X(f^*\N,\P)(U) = \Hom_{f^{-1}(U)}\left(f^*\N|_{f^{-1}(U)},\P|_{f^{-1}(U)}\right) \enspace \enspace \enspace (*) $$
but $f^*\N|_{f^{-1}(U)} = \left(f|_{f^{-1}(U)}\right)^*\N|_U$. Then by the adjunction, (*) is naturally isomorphic to
$$ \Hom_U\left(\N|_U, f_*\P|_U\right) = \HH_Y(\N,f_*\P)(U). $$
Naturality of this isomorphism tells us that this map commutes with the restrictions and so it is a map of sheaves and so we get an isomorphism.