$f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})\quad$? Let $f$ be a morphism of schemes $f: (X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$, and $\mathcal{F},\mathcal{G}$ be sheaves of $\mathcal{O}_Y$-modules. I am trying to prove (I do NOT claim this to be true):
$f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})$
By the definition of $f^{*}$, and the property of the tensor product, one can check that this boils down to proving: $\quad f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G} \cong f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G})$. However, I cannot continue this bare hand computation at the present stage. For one thing $f^{-1}$ and $\otimes$ both require sheafification, and thus I get a compostion of two sheafification objects; for another, I know nothing about good properties of stalks on $f^{-1}$.
I guess the computation may be  dirty, but I appreciate any insight on handling the problem.
 A: Alternative proof, using only adjunctions.
First, notice that there is an isomorphism in $\mathsf{Mod}(Y)$
$$f_* \underline{\hom}_X(f^* G,H) = \underline{\hom}_Y(G,f^* H)$$
for $G \in \mathsf{Mod}(Y)$ and $H \in \mathsf{Mod}(X)$. In fact, on an open subset $V \subseteq Y$, we have
$\Gamma(V,f_* \underline{\hom}_X(f^* G,H)) = \hom_{f^{-1}(V)}(f^* G |_{f^{-1}(V)},H|_{f^{-1}(V)})$
$ = \hom_{f^{-1}(V)}(f_V^* G|_V,H|_{f^{-1}(V)}) = \hom_V(G|_V,(f_V)_* H|_{f^{-1}(V)})$
$ = \hom_V(G|_V,(f_* H)|_V) = \Gamma(V,\underline{\hom}_Y(G,f^* H)).$
The rest is purely formal:
$\hom_X(f^* F \otimes f^* G , H)
 = \hom_X(f^* F , \underline{\hom}_X(f^* G,H))
 = \hom_Y(F,f_* \underline{\hom}_X(f^* G,H))$
$ = \hom_Y(F,\underline{\hom}_Y(G,f_* H)) = \hom_Y(F \otimes G,f_* H) = \hom_X(f^* (F \otimes G),H).$
Hence $f^* F \otimes f^* G \cong f^* (F \otimes G)$ by Yoneda. This proof also works in quite general contexts (for example where no stalks are available).
A: Yes, we have $f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})\quad$
And, yes, this results from the isomorphism $\alpha: f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G} \overset {\sim}{\longrightarrow} f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G})$
And, no, the computation is not dirty!
To prove that the natural map $\alpha$ is an isomorphism, it is enough to look at the stalks.
The  morphism  $\alpha_x: (f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G})_x \to (f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G}))_x$ is indeed an isomorphism because or the following two general results (which do not involve schemes):
Fact 1: For any continuous map $f:X\to Y$ of topological spaces and any sheaf $\mathcal E$ on $Y$, we have for every $x\in X$ a canonical isomorphism $(f^{-1} \mathcal E)_x=\mathcal E _{f(x)}$ 
Fact 2: Given a sheaf of rings $\mathcal A$ and sheaves $\mathcal C, \mathcal D$ of 
$\mathcal A$-Modules on the topological space $X$, we have for every $x\in X$ a natural isomorphism 
$(\mathcal C \otimes _{\mathcal A}\mathcal D)_x=\mathcal C_x \otimes _{\mathcal A_x}\mathcal D_x$.
[Of course, in the discussion at hand $\mathcal A$ is $f^{-1}\mathcal{O}_Y$]
