I want to show that the ses $0\rightarrow N \hookrightarrow M \twoheadrightarrow M/N \rightarrow 0$ splits when $M/N$ is free (they are all $R$-modules where $R$ denotes a ring). What I want to do is find a section $s:M/N\rightarrow M$ such that when I compose it with the quotient map $q:M\twoheadrightarrow M/N$ it gives me the identity on $M/N$.
I know this question has been asked multiple times on this forum, I just need some clarification. I do know that what I have to do is use the fact that I have a basis $(x_i+N)_{i\in I}$ for $M/N$ to construct my section. However, I have been thinking about it a lot and to me, the section defined by sending each element $x_i+N \rightarrowtail x_i$ would be enough, because then taking $(q\circ s)(x_i+N)=q(x_i)=x_i+N$. However, on this answer someone suggested to take the section that would do the following to the basis elements: $s(n,x_i+N)=n+x_i$. There is another answer to the same question here that does the same and I think that the purpose is to prove that the kernel of the section is trivial, but I do believe that the kernel of my section is already trivial. So then my question is, why are they taking the trouble of defining the section like that? Also, what exactly is the $n$ we are taking in the section?
You are defining section $M/N\to M$ and the links you provided are defining isomorphism $N\oplus M/N\to M$.