In Görtz' & Wedhorn's Algebraic Geometry, there is a proof on page 271 that I do not fully understand:
Theorem 10.83. Let $R$ be a noetherian integral domain, let $A$ be a finitely generated $R$-algebra, and let $M$ be a finitely generated $A$-module. Then there exists $s \in R \setminus \{0\}$ such that the localization $M_s$ is a free $R_s$-module.
Proof. Let $x_1, \cdots , x_n \in A$ be generators of $A$ as an $R$-algebra. We prove the proposition by induction on $n$. If $n = 0$, i. e. $A = R$, then we are done by Lemma 10.81. Otherwise let $A'$ be the $R$-subalgebra of $A$ generated by $x_1, . . . x_{n−1}$. Let $e_1, . . . , e_r$ be generators of $M$ as an $A$-module, and set $N = \sum A' e_i$. We can apply the previous lemma (10.82) to $A', A, M$ and $N$, and get a filtration of $M/N$ by finitely generated $A'$-submodules with only finitely many subquotients (up to isomorphism). Adding $N$ as a filtration step, we obtain a filtration of $M$ with the same property.
By the induction hypothesis, all the subquotients of this filtration are free over a suitable localization of $R$, and over this localization the filtration splits (???) (Proposition B.14). This shows the proposition.
The two results referenced in the proof are:
Lemma 10.82:
Prop. B.14:
Here's my question: why do these assumptions imply that the finite filtration
$$M=M_n ⊃ M_{n-1} ⊃ ... M_1 ⊃ M_0=N ⊃0$$
splits over suitable $R_s$? By induction hypothesis all quotients $M_i / M_{i-1}$ are free over $R_s$. What's the meaning of the formulation 'the filtration splits'?
Does it mean that for every inclusion $i: M_{i-1} \subset M_i$ we obtain a retract $r:M_i \to M_{i-1} $ with $id_{M_i}= r \circ i$? If yes, why does this make usage of induction hypothessis? Why is this enough to show the theorem?
Once you know that all the subquotients are free, patching everything together should be pretty quick.
Lemma: Let $R$ be a ring and $M$ an $R$-module with a composition series $0\subset M_1\subset M_2\subset\cdots\subset M_n=M$ so that $M_i/M_{i-1}$ is free for all $i$. Then $M$ is free.
Proof: First, we note that $M_1$ is free. Next, assume we've shown that $M_i$ is free - we'll now show that $M_{i+1}$ is free. Consider the exact sequence $0\to M_i\to M_{i+1}\to M_{i+1}/M_i\to 0$. As $M_{i+1}/M_i$ is free, it is projective, so by B.14.ii, there's a splitting $M_{i+1}/M_i\to M_{i+1}$, which is equivalent to a direct sum decomposition $M_{i+1}\cong M_i\oplus M_{i+1}/M_i$. As both $M_i$ and $M_{i+1}/M_i$ are free, this shows that $M_{i+1}$ is free. By induction, since $n$ is finite, eventually we show that $M$ is free. $\blacksquare$
The meaning of "the filtration splits" is exactly what you say. This doesn't use the induction hypothesis directly - the induction hypothesis is needed to guarantee $N_s$ is free for some $s\in R$ so that we may obtain a filtration with all subquotients free. Once we have that filtration over $R_s$, we just apply the lemma and the conclusion that $M_s$ is free follows.