A normal subgroup $K \trianglelefteq G$ is complemented if there is a subgroup $H \le G$ such that $H \cap K = \{e\}$ and $G = HK$, i.e., $G$ is the semidirect product of $H$ and $K$ (Wikipedia).
Let $G$ be a group with a free presentation $G \cong F / R$. Hopf's formula asserts that the Schur multiplier $M(G)$ is given by $M(G) \cong (R \cap [F, F]) / [R, F]$. When is $M(G)$ complemented in $R / [R, F]$?
Since $R \le F$, $R/[R,F]$ is an abelian group.
$$R/(R \cap [F,F]) \cong R[F,F]/[F,F] \le F/[F,F],$$
which is free abelian, so $R/(R \cap [F,F])$ is also free abelian, and hence $(R \cap [F,F])/[R,F]$ has a complement in $R/[R,F]$.