Is there a functor $F$ of left modules preserving $\oplus$ by arbitrary isomorphism, but its restriction to fin. gen. proj. modules isn't additive? Are there rings $R$, $S$ and a functor $F:{_R\textbf{Mod}}\to{_S\textbf{Mod}}$ such that


*

*For all left $R$-modules $M,N$, we have $F(M\oplus N)\cong F(M)\oplus F(N)$
via an arbitrary isomorphism, and such that $F(0)=0$

*$F(R)$ is a finitely generated, projective left $S$-module,


but when we restrict and corestrict $F$ to the full subcategory of finitely generated projective left $R$- resp. $S$-modules, which we denote by $\mathcal{P}(R)$ resp. $\mathcal{P}(S)$, then this restricted functor $F:\mathcal{P}(R)\to\mathcal{P}(S)$ isn't additive, in the sense that it doesn't preserve split exact sequences?
With the help of this forum, I found out that when $S$ is stably finite (i.e whenever $M\oplus S^n\cong S^n$ for a left $S$-module $M$ and a $n\geq 0$, then $M=0$), then such a functor $F$ as above cannot exist. The proof of this is very similar to the answer of Jeremy Rickard in this post, and the condition of $S$ being stably finite is exactly what makes the proof work in this more general situation. Also note that if $S$ is commutative then it is stably finite.
One idea I had was to use the functor $F$ constructed in this answer, and give the abelian group $FA$ somehow naturally the structure of a e.g. $\operatorname{End}_{\mathbb{Z}}(\mathbb{Z}^{\oplus \mathbb{N}})$ module, such that $F\mathbb{Z}$ becomes finitely generated and free, but maybe this approach is too optimistic.

Let $R$ be a field $k$, $S=\operatorname{End}_k(k^{\oplus\mathbb{N}})$, and $F(M)=S\otimes_k (M\otimes_k M)$.  Note that $S$ has the property that $S\cong S^n$ as an $S$-module for all finite $n>0$.  If $M$ is nontrivial and finite-dimensional, then so is $M\otimes_k M$, so $F(M)\cong S$.  If $M$ is infinite-dimensional, then $M\otimes_k M$ has the same dimension, so $F(M)$ is free over $S$ of rank $\dim M$.  It follows that $F(M\oplus N)\cong F(M)\oplus F(N)$ for any $M$ and $N$.  However, $F$ is not additive, even when restricted to finite-dimensional vector spaces.
