Exactness of sequences of modules is a local property
Yes, exactness is indeed local, and localisation commutes with $\ker$ and $\operatorname{im}$ (since localisation is exact). In fact, exactness is so local that you just need to check it at the maximal ideals. Here is a sketch:
$M = 0$ if and only if $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$.
A homomorphism $M \to N$ is a monomorphism/epimorphism/isomorphism if and only if $M_\mathfrak{m} \to N_\mathfrak{m}$ is a monomorphism/epimorphism/isomorphism for all maximal ideals $\mathfrak{m}$. [Use (1).]
Suppose we have a sequence of modules and homomorphisms: $$0 \longrightarrow M'' \longrightarrow M \longrightarrow M' \longrightarrow 0$$ Suppose also that this sequence is exact after localising at $\mathfrak{m}$, for all maximal ideals $\mathfrak{m}$. Then, by (1), the sequence is a chain complex, and by (2), the sequence is exact at $M''$ and $M'$. Since we have a chain complex, there is an induced homomorphism $\ker (M \to M') \to \operatorname{coker} (M'' \to M)$; but this is an isomorphism after localising at each $\mathfrak{m}$, so the homomorphism is already an isomorphism, and thus the sequence is exact at $M$ as well.
As this is about one of the only things I can comment on I thought I'd write an answer! Hopefully it will be of help to someone. It essentially the same as what has been written but a bit more condensed.
$(1)$ $M = 0$ if and only if $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$.
Keep notation and hypothesis of the original post i.e. we are considering a sequence $E$: $M'\xrightarrow{f} M\xrightarrow{g} M''$ that is exact when localized at every maximal ideal.
Since $((g\circ f)M')_{\mathfrak{m}}=(g_\mathfrak{m}\circ f_\mathfrak{m})M_{\mathfrak{m}}'=0$, we have by $(1)$ that $E$ is a complex, and hence, the quotient module $\ker(g)/\operatorname{im}(f)$ is well-defined. Whence, $(\ker(g)/\operatorname{im}(f))_\mathfrak{m}=\ker(g_{\mathfrak{m}})/\operatorname{im}(f_{\mathfrak{m}})=0$, and the result follows.