Center of a quotient of a free group

I encountered the following statement while going through an old result by C.F.Miller (Ref. The Schur Multiplier, Gregory Karpilovsky, Theorem 2.6.6, Page 72):

Let $G=F/R$ be a presentation of an arbitrary group (i.e., $F$ is a free group and $R$ is a normal subgroup of $F$). Then $R/[F,R]$ is precisely equal to the center of $F/[F,R]$.

It is obvious that $R/[F,R]$ is a central subgroup of $F/[F,R]$. The converse doesn't seem so easy for me: for any $x \in F$ with $[F, x] \subseteq [F,R]$, then we should have $x \in R$. Taking $F = Z$ gives a counterexample for sure. However, I don't see any easy counterexample for any free group $F$ with rank at least two.

Any finite presentation of a finite cyclic group, like $G = \langle a,b \mid a^2,b \rangle$ gives the same contradiction.

$G \cong F/R \cong \frac{F/[F,R]}{R/[F,R]}$ is cyclic with $R/F,R] \le Z(F/[F,R])$, so $F/[F,R]$ is abelian.

But there are other finite groups, like $Q_8$ (see here) that cannot occur as $G/Z(G)$, so they give rise to further counterexamples, such as $\langle a,b \mid a^2=b^2, a^{-1}ba=b^{3} \rangle$.