Linearity of the right inverse of a surjective linear map

If you write $V=\ker f\oplus W $ (which can always be done), then the restriction $f:W\to V $ is bijective, and you can use its inverse as your linear right-inverse.

^{Since you're assuming that surjective maps have a right inverse (which is equivalent to the axiom of choice), you know that every vector space has a basis.}

Let $f\colon V\to W$ be a surjective linear map. Let $\mathscr{B}=\{w_\alpha:\alpha\in A\}$ be a basis of $W$; for each $\alpha\in A$, pick $v_\alpha$ such that $f(v_\alpha)=w_\alpha$ and define $g\colon W\to V$ by stipulating that $$ \text{for each $\alpha\in A$, $g(w_\alpha)=v_\alpha$} $$ By standard results, $g$ is well defined; moreover $f\circ g$ is the identity on $\mathscr{B}$, so it is the identity on $W$.