Let $\phi:R\to S$ be a flat ring homomorphism and consider the induced adjoint pair
$$\phi\_!:R-Mod\rightleftarrows S-Mod:\phi^\*,$$
where $\phi\_!=(S\otimes\_R -)$. The right adjoint $\phi^\*$ is easily described if we see $\phi$ as an additive functor between the one-object categories $R$ and $S$ and we view a left $S$-module $M$ as an additive functor $M:S\to Ab$. In this case, $\phi^\*$ is just composition by $\phi$, $(M:S\to Ab)\mapsto (M\circ\phi:R\to Ab)$.

By definition of flat homomorphism, $\phi\_!$ is an exact functor, while $\phi^\*$ is exact for any ring homomorphism $\phi$.

Consider now the categories of finitely presented modules $fp(R)$ and $fp(S)$, and restrict the functor $\phi\_!$ to an additive functor $F:fp(R)\to fp(S)$. Denote also by $fp(R)-Mod$ and $fp(S)-Mod$ the categories of additive functors $fp(R)\to Ab$ and $fp(S)\to Ab$, respectively. Of course, also here there is an induced exact functor 
$$F^\*:fp(S)-Mod\to fp(R)-Mod$$
and, using additive Kan extensions one may construct a left adjoint 
$$F\_!:fp(R)-Mod\to fp(S)-Mod.$$
Can we prove that the functor $F\_!$ is exact?

Spelling this out, rember that, given a sequence of left $fp(R)$-modules
$$0\to M\_1\to M\_2\to M\_3\to 0$$
this sequence is exact if and only if, for all $P\in fp(R)$, the sequence of abelian groups
$$0\to M\_1(P)\to M\_2(P)\to M\_3(P)\to 0$$
is short exact. Thus one should prove (or disprove) that for any short exact sequence as above, 
$$0\to F\_!M\_1(Q)\to F\_!M\_2(Q)\to F\_!M\_3(Q)\to 0$$ 
is short exact for all $Q\in fp(S)$.