I think one

One can show that $\cal G$ is a direct sum of locally free coherent sheaves, which is not quite what you want, but almostprove this also without Bass's theorem. Let $X= Spec A$ and $Y=Spec B$. Now The sheaf $\cal F$ comes from a finitely generated module $M$ over $C=A\otimes B$. 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 $M\oplus N=\bigoplus_{i=1}^kC e_i$. 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 $(b_j)_{j\in J}$ be a basis of $B$ over the ground field, then $$M=\bigoplus_{j\in J}b_j M_0.$$ Let $N_0$ be the $A$-module generated by $n_1,\dots,n_k$. Then $M_0\oplus N_0= \bigoplus_i Ae_i$ is a free $A$-module and so is Also $b_j(M_0\oplus N_0)=\bigoplus_iAb_je_i$. 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.

1

I think one can show that $\cal G$ is a direct sum of locally free coherent sheaves, which is not quite what you want, but almost. Let $X= Spec A$ and $Y=Spec B$. Now $\cal F$ comes from a finitely generated module $M$ over $C=A\otimes B$. Our 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 $M\oplus N=\bigoplus_{i=1}^kC e_i$. 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 $(b_j)_{j\in J}$ be a basis of $B$ over the ground field, then $$M=\bigoplus_{j\in J}b_j M_0.$$ Let $N_0$ be the $A$-module generated by $n_1,\dots,n_k$. Then $M_0\oplus N_0= \bigoplus_i Ae_i$ is a free $A$-module and so is Also $b_j(M_0\oplus N_0)=\bigoplus_iAb_je_i$. This means that we have written $M$ as a direct sum of finitely generated projective $A$-modules as claimed.