An elementary lemma of commutative algebra

Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $R$-modules, with $R$ Noetherian and $C$ of finite type, giving rise to a class $c \in Ext^1_R(C,A)$. Tensoring with $R_{\mathfrak m}$ for a maximal ideal $\mathfrak m$ gives the short exact sequence $0 \to A_{\mathfrak m} \to B_{\mathfrak m} \to C_{\mathfrak m} \to 0$, which corresponds to the image of $c$ in the $Ext^1_{R_{\mathfrak m}}(C_{\mathfrak m},A_{\mathfrak m}) = R_{\mathfrak m}\otimes_R Ext^1_R(C,A)$. (This isomorphism holds because $R_{\mathfrak m}$ is flat over $R$; to prove it one computes the Ext via a resolution of $C$ by finite rank free $R$-modules, which exists since $R$ is Noetherian and $C$ is of finite type.)

Now suppose that these localizations are all split. Then we have an element $c$ in the $R$-module $Ext^1_R(C,A)$ whose image in the localization at each maximal ideal $\mathfrak m$ vanishes. It follows that this element itself vanishes. (Its annihilator is not contained in any maximal ideal.)

So the answer seems to be "yes": for Noetherian $R$, if a short exact sequence whose third term is finite type splits locally at every $\mathfrak m$, it splits.

Note: I think one can also see this by comparing sheaf Exts with actual Exts for the associated quasi-coherent sheaves on Spec $R$. The point is that there is a spectral sequence involving the cohohomology of the sheaf Exts converging to the actual Exts (in some generality), but on the affine Spec $R$ the higher cohomology of the sheaf Exts vanishes, so that a class in any $Ext^i$ is a global section of the sheaf Ext, and hence is determined by what happens locally. (Hopefully this is not all nonsense.)

If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof):

Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a short exact sequence of $R-$modules with $C$ finitely presented and assume it splits after localisation at every maximal ideal.

Use the natural isomorphism $\operatorname{Hom}_{R_m}(C_m,A_m)=R_m\otimes\operatorname{Hom}_R(C,A)$ (which uses the flatness of localisation and the fact that $C$ is finitely presented) and the assumption to see that the map $\operatorname{Hom}_R(C,B)\rightarrow\operatorname{Hom}_R(C,C)$ is surjective since all of its localisations are.

Since the giving a splitting is equivalent to this map being surjective we are done.

P.S.: Of course one could prove "if and only if" in the statement like this.

Counterexample in the general case: Take $R=\prod_{\mathbb N} \mathbb F_2$, $I=\sum_{\mathbb N} \mathbb F_2$ and $C$ the cokernel of the inclusion.

Now I claim three things:

1.) $R$ has dimension zero

Proof: Every element of $R$ is an idempotent and every prime ideal in $R$ has to contain exactly one of $e$ or $1-e$ for every idempotent $e$ in R. If we had a chain of prime ideals, the larger one would necessarily have to contain $e$ and $1-e$ for one such idempotent and so couldn't exist.

2.) The localisation of $R$ at every maximal ideal is a field.

Proof: It is a zero-dimensional local ring by 1.) and reduced since $R$ doesn't contain nilpotent elements. Thus this localisation is a field.

3.) $I$ is not a direct summand of $R$.

Proof: Direct summands correspond to idempotents in $R$. Since every element in $R$ is idempotent we need to analyse all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be $I$. If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements. Since $I$ contains countably many elements, it can't be a direct summand of $R$.

This establishes the counterexample, since the short exact sequence $0\rightarrow I\rightarrow R\rightarrow C\rightarrow 0$ splits in every localisation at a maximal ideal.