Assume $X\longrightarrow S$ is a nice scheme, and $\mathcal{F}$ is a sheaf on $X$. Let $\pi_i: X\times_S X\longrightarrow X$ be the canonical projections, $i\in\{1,2\}$.
I read in a book of Berthelot and Ogus that the sheaf $(\pi_1)_*(\pi_2)^*\mathcal{F}$ is isomorphic to the sheaf $\mathcal{O}_X\otimes_{\mathcal{O}_S}\mathcal{O}_X\otimes_{\mathcal{O}_X}\mathcal{F}$ when $X$ is affine.
Why should this not hold generally when $X$ is not affine?
Why do we need to take pullback via one projection and pushforward via another projection? Wouldn't we get the same sheaf if we consider $(\pi_1)_*(\pi_1)^*$? What is the point of that operation?
Following up on one of my comments. Suppose $X:=Spec(A)$ is affine and $S:=Spec(B)$ is affine and so $A$ is just a $B$-algebra. Then it is known e.g. Hartshorne Chapter II.5 that $f^*\mathscr{M}=\widetilde{M\otimes_B A}$ for $\mathscr{M}=\widetilde{M}$ a coherent $\mathcal{O}_S$-module. Whereeas $f_*\mathscr{N}=\widetilde{N|_B}$ for $\mathscr{N}=\widetilde{N}$ a coherent $\mathcal{O}_X$-module (coherence isn't important here).
Onto the matter at hand. Let me answer your point 2. first.
Let $\mathscr{F}=\widetilde{M}$ be a coherent sheaf on $X$ and $M$ an $A$-module. Then the claim boils down to saying that $(M\otimes_A (A\otimes_B A))|_A$ is isomorphic to $M\otimes_A (A\otimes_B A)$ as an $A$-module. This is certainly clear I hope.
Now what if you swapped it with $\pi_{1*}\pi_1^*$? Then you'd be saying $(M\otimes_A(A\otimes_B A)|_A$ is isomorphic to $M\otimes_A (A\otimes_B A)$. What's the difference because they look the same?
The matter is how you choose to take the restriction $|_A$ !! The restriction in the first case (1) comes from restricting via the $A$-module structure given by $A\rightarrow A\otimes_B A$ sending $a\mapsto a\otimes 1$ while the second one (2) swaps $a\mapsto 1\otimes a$.
But how was the tensor product taken? The extension of scalars was done by viewing $A\rightarrow A\otimes_B A$ as an $A$-module via $a\mapsto 1\otimes a$ in both cases. So one can check that only the first choice (1) gives the right isomorphism.
Warning. This part of the answer is vague. Now let me explain the issue with 1 from what I understand. If $X$ were not affine, we would at least hope for quasicoherent $\mathscr{F}$ for something like this to hold. Even then, it is not readily clear to me that affine is absolutely needed here. For pushforwards to be well-behaved w.r.t. quasioherence, we'd need hypotheses on the maps such as in Hartshorne Proposition II.5.8. But the projection map need not satisfy quasicompactness e.g. $\mathbb{A}^\infty\times\mathbb{A}^\infty\rightarrow \mathbb{A}^\infty$ is not quasicompact.