Let $M,M',M''$ $R$-modules, $R$ commutative ring with unit. $$0\xrightarrow{n_1} M'\xrightarrow{f} M \xrightarrow{g} M''\xrightarrow{n_2} 0$$ It is exact if and only if $M'\subseteq M$ submodule and $M'' \cong M/M'$.
The Proof of $\Rightarrow$ is clear. Indeed $\newcommand{\Im} {\operatorname{Im}} \Im(n_1) ={0}=\ker(f) \Rightarrow f$ is injective, then $f:M'\to M$ is an isomorphism and then $M'\cong \Im(f)$. So as we know from exactness that $ \Im(f) = \ker(g)$ we have $\ker(g)\cong M'$ and from the first theorem of isomorphism we have $\Im(g)\cong M/M'$. But $\ker(n_2) = M'' = \Im(g)$ and we have done.
What about the converse instead?