Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Suppose that $\text{Hom}_R(M'',{-})$ and $\text{Hom}_R(M',{-})$ and commute with filtered colimits. This should be enough to show that $\text{Hom}_R(M,{-})$ commutes with filtered colimits.
Take some filtered colimit $Q=\text{colim} Q_i$. The Hom functor is left exact, so $$0 \to \text{Hom}_R(M'',Q) \to \text{Hom}_R(M,Q) \to \text{Hom}_R(M',Q)$$ is exact. There are obvious comparison maps from this to $$\text{colim }\text{Hom}_R(M'',Q_i) \to \text{colim }\text{Hom}_R(M,Q_i) \to \text{colim }\text{Hom}_R(M',Q_i).$$ By assumption the outer two comparison maps are isomorphisms.
How can we show that the desired middle map is an isomorphism? It would be enough to show that $\text{Ext}^1_R(M'',{-})$ commutes with filtered colimits, but I don't see why this is true.
Context: For an $R$-module $M$, being finitely presented is equivalent to $\text{Hom}_R(M,{-})$ commuting with filtered colimits. It is certainly true that if $M'$ and $M''$ are finitely presented then $M$ is. Together this gives the result. However, I want to see how this argument should go working directly with the filtered colimit definition.