Let $X$ and $Y$ be schemes and let $F$ be a sheaf on $Y$. Let $f: X \rightarrow Y$ be a morphisme of schemes. Define the inverse image of $F$, $$f^*F:= f^{-1}F\otimes_{f^{-1}\mathcal{O}Y}\mathcal{O}_X$$
For an open subset $U\subseteq X$, $f$ be the inclusion $U\hookrightarrow X$ and $F$ be a sheaf on $X$. Then is it true that $f^*F=F|_{U}$?
It is, since in this case the structural map $f^{-1} \mathcal{O} _{X} \rightarrow \mathcal{O} _{U}$ is an isomorphism, because the functors $f^{-1}(-)$ and $(-) |_{U}$ coincide (essentially by definition) and $\mathcal{O} _{U} = \mathcal{O} _{X} |_{U}$. So you actually have
$f^{*} \mathcal{F} = f^{-1}\mathcal{F} \otimes _{f^{-1} \mathcal{O} _{X}} \mathcal{O} _{U} \simeq f^{-1}\mathcal{F} = \mathcal{F} |_{U}$
The question doesn't make any sense for two reasons: The definition of $f^* F$ as a tensor product makes only sense when $F$ is a sheaf of modules on $Y$, not just a sheaf. And $f^* F|_U$ doesn't make much sense. Perhaps you mean $f^* F$ and you want to prove that $f^* F \cong F|_U$? Of course it is not important that $X,Y$ are schemes, all this works for arbitrary ringed spaces.
The most important property of $f^*$ (and in fact, from a more abstract point of view, this is the definition) is that it is left adjoint to $f_*$ (both for categories of sheaves, as well as for categories of sheaves of modules). If $f$ is the inclusion of an open subspace $U \hookrightarrow X$, then one can verify directly that $(-)|_U$ is left adjoint to $f_*$, since there is a canonical bijection $\hom(F|_U,G) \cong \hom(F,f_* G)$ given by $\alpha \mapsto (V \mapsto \alpha(V \cap U))$. It follows that $f^* \cong (-)|_U$.
The explicit construction of $f^*$ is not really important, you can always derive anything from its universal property.
