We know that the tensor functor in modules preserves right exactness, that is, if $$M' \rightarrow M \overset{f}{\rightarrow} M''\rightarrow 0$$ exact then $$F\otimes M' \rightarrow F\otimes M \rightarrow F\otimes M''\rightarrow 0$$ is exact. Moreover tensoring by flat modules preserves also left exactness. In all proofs for this result (that I've ever seen!), the surjectivity of $f$ is used. What we can say about the exact sequences in which the right map is not surjective? For example if $$0\rightarrow M' \rightarrow M \overset{f}{\rightarrow} M''$$ exact sequence of $R-$modules ($M, M'$ and $M''$ are not zero and $f$ is not epimorphism) and $F$ is a flat $R-$module, then is the induced sequence $$0\rightarrow F\otimes M' \rightarrow F\otimes M \rightarrow F\otimes M''$$ exact? If it is not true, what conditions on $R$ or $F$ are needed?
I tried to find a direct proof, but it seems difficult, because of different representations of any element in tensor products. I know that we can always change $M''$ to $f(M)$ to have $f$ as an epimorphism. But I was wondering if doing this for $M''$ makes difficulties in some problems.
I will be appreciated if anyone can help.