Every $R$-module is a quotient of a free module I'm reading through my notes, and I have a theorem as the title says. As proof I have only a picture.

where $v:m\mapsto(0,...,0,1,0,...,0)$ and 1 is at $m$-th position.
I know, that $\oplus_M R$ is a free module generated by set  $M$, but is this all? I do not see how the statement is proved.

You can think to the free module $F$ generated by $M$ as the set of functions $f\colon M\to R$ such that $\{m\in M:f(m)\ne0\}$ is finite. The map $M\to F$ is $x\mapsto e_x$, where
$$
e_x(m)=\begin{cases} 1 & m=x \\[4px] 0 & m\ne x \end{cases}
$$
Addition and scalar multiplication on $F$ are obviously defined. The surjective homomorphism $\varphi\colon F\to M$ is given by
$$
\varphi(f) = \sum_{m\in M}f(m)m
$$
Clearly,
$$
\varphi(e_x)=x
$$
