The tensor product $F \otimes G$ has the following universal property:

$\hom(F \otimes G,H) = \text{Bilin}(F \times G,H)$

Here, the right hand side is the set of bilinear sheaf homomorphisms $F \times G \to H$. You should keep this in mind instead of the explicit construction involving sheavifications! The idea behind tensor products is not their construction, but rather that they classify bilinear maps. This idea is familiar from commutative algebra and should not be lost in algebraic geometry.

Now we also have a canonical identification

$\text{Bilin}(F \times G,H) = \hom(F,\underline{\hom}(G,H))$.

The pullback $f^\*$ has also a universal property, it is left adjoint to the pushforward $f_\*$. Thus, we get:

$\hom(f^\* F \otimes f^\* G,-) \cong \hom(f^\* F,\underline{\hom}(f^\* G,-)) \cong \hom(F,f_* \underline{\hom}(f^\* G,-))$
$\cong \hom(F,\underline{\hom}(G,f_\* -)) \cong \hom(F \otimes G,f_\* -) \cong \hom(f^\* (F \otimes G),-)$

and the Yoneda lemma gives us $f^\* F \otimes f^\* G \cong f^\* (F \otimes G)$.

PS: Try to prove $F \otimes (G \otimes H) \cong (F \otimes G) \otimes H$ using sheavifications and compare the proof with the one using trilinear maps.