Let $R$ be a ring with identity and $M$ be an $R$-right module. Then $M$ is called free over $X \subseteq M$ if for every module $N$ with mapping $\alpha : X \to N$ we can extend it uniquely to a homomorphism $\overline \alpha : M \to N$. For a ring $R$ denote by $R_R$ the module structure if $R$ is regarded as a module over itself. Now I am looking for a slick and short proof of the following
A right $R$-module $M$ is free (over some set $X$) if and only if it is a direct sum of copies of $R_R$.
By some "high-level" theorems, like that $X$ is a minimal generating set for $M$ and the elements of $X$ are lineary independent in $M$ (i.e. every element from $M$ could be uniquely written as linear combination of finitely many elements from $X$) we get an isomorphism by mapping onto the "coordinates" and the direct product of a ring is module isomorphic to the direct sum of $R_R$, thus showing that every free modules is isomorphic to such a direct sum. But I hoped there might be some easier argument, without supposing linear independence of the $X$ and that $M = \langle X \rangle$.
My attempt. If $x \in X$ consider the $R$-right module $xR$ and the mapping $\beta : X \to \{x\}$, then we have a unique $\overline \beta : M \to xR$ which happens to be surjective as $\overline \beta(xr) = \beta(xr) = xr$. Now $M = xR \oplus \mbox{ker}(\overline \beta)$ as $m = rx + (m - rx)$ with $rx = \overline\beta(m)$. Now I wanted to show that $\mbox{ker}(\overline \beta)$ is free over $X \setminus \{x\}$ and hence proceed inductively. But then I realised that this is just possible if $X$ is countable or has some other structure which permits such inductive reasoning. So I guess such a reasoning is not valid in general, right? If this proof scheme does not work, are there any other short proofs of this fact?
It's not entirely clear to me what you are looking for exactly, but here is a short proof of your statement:
Consider the $R$-module $N= \bigoplus_{x \in X} R_x$, where each $R_x$ is just a copy of $R_R$. Denote the unit of $R_x$ by $1_x$. Then the map of sets $f: X \to N, x \mapsto 1_x$ gives a unique module homomorphism $M \to N$ extending $f$ (and which I will abusively denote by $f$ as well). Furthermore, the maps $g_x: R \to M, g_x(r) = xr$ sum up to give a homomorphism $g:N \to M$. I claim that $f \circ g = \mathrm{id}_N$ and $g \circ f = \mathrm{id}_M$. For the first equality, consider an element $a = (a_x)_{x \in X} \in N$. Then $f \circ g(a) = f(\sum_x x a_x) = \sum_x f(x) a_x = \sum_x a_x = a$ which proves the first identity. For the latter equality, we have $g \circ f (x) = g_x(1_x) = x$, and hence $g \circ f$ is the identity on $X$. Since M is free, and $g \circ f$ is a module homomorphism, this means that it must be equal to $\mathrm{id}_M$.