"PushPull" of a locally free sheaf Let $F \to Y$ be a locally free sheaf over $X$, and $\pi:X\to Y$. Is it true that
$$ \pi_*\pi^{-1}F = F \quad ? $$
This seems obvious to me, but I'm asking here because I'm often wrong with these things ;-)
My way to prove the above claim is simply to consider an affine open $V\subset Y$ and compute
$$ \pi_*\pi^{-1}F(V) = \pi^{-1}F(\pi^{-1}V) = \lim_{W\supset V}F(W) = F(V) $$
This really looks correct to me. In case it is wrong, could you point out which is the step that goes wrong?
Bonus question: What about the case of a quasichoerent sheaf on a scheme? Is it true that $$\pi_* \pi^* F = F \quad ?$$
 A: The second formula you quote is true in the following case:

$Z \overset{\pi}\to X$ is a closed immersion with $Z\,$ equal to the support of a coherent sheaf $\mathcal{F}$ on $X$. 

Here is a proof of this fact. We may assume that $X$ is affine (equal to $\operatorname{Spec} A$) with $\mathcal{F} = \widetilde{M}$ for some finite type $A$-module $M$. Hence $\pi$ is the closed immersion $\operatorname{Spec}(A/\operatorname{Ann}(M)) \to \operatorname{Spec} A$.
Now we have  $\pi_{\ast} \pi^\ast \widetilde{M} = \hspace{1mm}_A(M \otimes_A A/\operatorname{Ann}(M))$ which is canonically isomorphic to $M$ as an $A$-module because $M$ is canonically an $A/\operatorname{Ann}(M)$-module. This concludes the proof.
A: Neither is true.  In the first your mistake is in this step:
$$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset V}F(W)$$
It should be
$$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset \pi\pi^{-1}V}F(W)$$
because in general we don't have $\pi(\pi^{-1}(V)) = V$.  Take, for example, $\pi$ to be the inclusion map of a point.
For the second assume $X$ and $Y$ are affine varieties over a field $k$.  Then quasicoherent sheaves correspond to modules and $\pi^*$ and $\pi_\ast$ correspond to tensor and restriction respectively.  If $A \to B$ is a ring map and $M$ an $A$-module then $B \otimes_A M$ is not necessarily isomorphic to $M$ as an $A$-module.  Again, choosing the inclusion of a point provides a counterexample.  This corresponds to a ring map $A \to k$ so $A/\mathfrak m \simeq k$ where $\mathfrak m$ is the kernel of the inclusion.  Then $M \otimes k \simeq M/\mathfrak mM$ and if $\mathfrak mM \neq 0$ then this is not isomorphic to $M$ as an $A$-module.
