ag.algebraic geometry - Does a locally free sheaf over a product pushforward to a locally free sheaf?

One can prove this also without Bass's theorem.
Let $X= Spec A$ and $Y=Spec B$.
The sheaf $cal F$ comes from a finitely generated module $M$ over $C=Aotimes B$.
Our first goal is to show that $M$, as an $A$-module, is a direct sum of finitely generated, locally free modules.
Since $A$ is noetherian, this is equivalent to $M$ being a direct sum of finitely generated projective $A$-modules.

The fact that $M$ is locally-free implies that $M$ is projective over $C$, further it is finitely generated, so there is a finitely generated $C$-module $N$ such that $Moplus N=bigoplus_{i=1}^kC e_i$.
Now each $e_i$ of this free basis can bewritten uniquely as $e_i=m_i+n_i$.
Let $M_0$ be the $A$-module generated by $m_1,dots,m_k$.
Let $(b_j)_{jin J}$ be a basis of $B$ over the ground field, then
$$
M=bigoplus_{jin J}b_j M_0.
$$
Let $N_0$ be the $A$-module generated by $n_1,dots,n_k$. Then $M_0oplus N_0= bigoplus_i Ae_i$ is a free $A$-module and so is
Also $b_j(M_0oplus N_0)=bigoplus_iAb_je_i$.
This means that we have written $M$ as a direct sum of finitely generated projective $A$-modules as claimed.

Now to conclude remember that $A$ is noetherian, therefore for each point in $X$ there exists an open neighborhood, where all summands of $cal G$ are free.