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 $Q\in fp(S)$.



Exactness of an additive left Kan extension
This is true, even if the ring extension is not necessarily flat. It follows from the fact that $F$ is right exact.


First some generalities: given an additive category $\mathcal{A}$ I will write $\mathcal{PA}$ for the category of additive presheaves (that is, additive functors ${\mathcal{A}}^{\mathrm{op}} \rightarrow \mathrm{Ab}$). The question when additive left Kan extensions of additive functors with target $\mathrm{Ab}$ along the Yoneda embedding $Y \colon \mathcal{A} \rightarrow \mathcal{PA}$ are exact is well studied: given an additive functor $F \colon \mathcal{A} \rightarrow \mathrm{Ab}$, the left Kan extension $\mathrm{Lan}\_Y F \colon \mathcal{PA} \rightarrow \mathrm{Ab}$ is exact if and only if $F$ is *flat*, which by definition means that the category of elements of $F$ is cofiltered. This follows in particular if $\mathcal{A}$ has finite limits and $F$ preserves them.


The theory of flat functors at this level of generality is for example developed in the paper by Oberst and Röhrl "Flat and coherent functors," Journal of Algebra, Volume 14.


If we have a functor $F \colon \mathcal{A} \rightarrow \mathcal{B}$ we can use this to study exactness of the left Kan extension $\mathrm{Lan}\_{Y} YF \colon \mathcal{PA} \rightarrow \mathcal{PB}$. (For the case we're interested in, the functor $F$ here will be the *opposite* of the functor $F$ in the question.) This functor is exact if and only if its composite with all evaluation functors $\mathrm{ev}\_B \colon \mathcal{PB} \rightarrow \mathrm{Ab}$ is exact. As a left adjoint, $\mathrm{ev}\_B$ preserves left Kan extensions. This reduces the problem to checking that for all objects $B \in \mathcal{B}$, the functor $\mathcal{B}(B,F-) \colon \mathcal{A} \rightarrow \mathrm{Ab}$ is flat. This is in particular the case if $\mathcal{A}$ has finite limits and $F$ preserves them.


To apply this in the situation at hand, we take $\mathcal{A}=fp(R)^{\mathrm{op}}$ and $\mathcal{B}=fp(S)^{\mathrm{op}}$. Both these categories have finite limits (since finite colimits of finitely presentable modules are again finitely presentable), and the functor $\mathcal{A} \rightarrow \mathcal{B}$ preserves finite limits (since $\phi\_!$ preserves finite colimits). From the result mentioned above it follows that the induced functor $\mathcal{PA} \rightarrow \mathcal{PB}$ is exact. 


Note that we did not use the fact that $\phi\_!$ preserves finite limits as well. This would be necessary to prove exactness of the induced left adjoint between the categories of *contravariant* functors on $fp(R)$ and $fp(S)$. In fact, here we run into the difficulty that these categories do not in general possess finite limits, so we have to go back to the definition of flatness to prove this "dual" result.


Edit: here are some details for the dual result (where $\mathcal{A}=fp(R)$ and $\mathcal{B}=fp(B)$), which is indeed a bit trickier. As mentioned above, we need to check that the category of elements of $\mathcal{B}(B,F-)$ is cofiltered for every finitely presentable $S$-module $B$. It is certainly non-empty since it contains the zero morphism. Moreover, given two objects $(A, f \colon B \rightarrow FA)$ and $(A^{\prime},f^{\prime} \colon B\rightarrow FA^{\prime})$, the two projections out of the direct sum give morphisms from $(A\oplus A^{\prime}, (f,f^{\prime}) \colon B \rightarrow F(A\oplus A^{\prime}))$ to these two objects in the category of elements. It only remains to check the condition on a pair of morphisms between two objects in the category of elements. Since $F$ is addtitive, this reduces to the following condition: Given a morphism $f \colon B \rightarrow FA$ of $S$-modules and a morphism $g \colon A \rightarrow A^{\prime}$ between finitely presentable $R$-modules with $Fg \circ f=0$, there exists a finitely presentable $R$-module $K$, morphisms $k \colon K \rightarrow A$ and $h \colon B \rightarrow FK$ such that $gk=0$ and $Fk \circ h=g$.


To see that such a pair $(K,k)$ exists, let $k^{\prime} \colon K^{\prime} \rightarrow A$ be the kernel of $g$ in the category of *all* $R$-modules. By assumption, the functor $\phi\_!$ preserves this kernel, so we get a morphism $f^{\prime} \colon B \rightarrow \phi\_! K^{\prime}$ such that $\phi\_! k^{\prime} \circ f^{\prime}=f$. We would be done if $K^{\prime}$ were finitely presented, which of course won't be the case in general. However, we can always write $K^{\prime}$ as filtered colimit $k\_i \colon K\_i \rightarrow K^{\prime}$ of finitely presentable $R$-modules $K\_i$. This filtered colimit is preserved by the left adjoint $\phi\_!$. Moreover, since $B$ is a finitely presentable $S$-module, there exists an index $i\_0$ such that the morphism $f^{\prime} \colon B \rightarrow \phi\_! K^{\prime}$ factors through $\phi\_!(k\_{i\_0}) \colon \phi\_! K\_{i\_0} \rightarrow \phi\_! K^{\prime}$, that is, there exists a morphism $g \colon B \rightarrow \phi\_! K\_{i\_0}$ such that $\phi\_! (k\_{i\_0}) g=f^{\prime}$. The object $K=K\_{i\_0}$ and the morphisms $k=k^{\prime} \circ k\_{i\_0}$ and $g$ thus have the desired properties.


