If a filtration of a module $M$ splits, then $M$ is locally free (Görtz & Wedhorn) 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?
 A: 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.
