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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 nescessarily 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 analyze 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→I→R→C→0 splits in every localisation at a maximal ideal.

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

Let 0→A→B→C→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 Hom$_{R_m}(C_m,A_m)$=R$_m$⊗Hom$_R$(C,A) (which uses the flatness of localisation and the finitely presentedness of C) and the assumption to see that the map Hom$_R$(C,B)→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.