@DerekHolt 's simple elegant answer makes this answer redundant, but for variety and a visual argument:
Let $F_2$ be freely generated by elements, $a,b$ and let $A=F_2/F_2'\cong \mathbb{Z}\oplus \mathbb{Z}$. Then $F_2'$ is freely generated by elements $\{e_x\}_{x\in A}$ where $e_{(i,j)}=a^ib^j[a,b]b^{-j}a^{-j}$.
Let $\mathbb{R}^A$ denote the real vector space with basis elements $\{v_x\}_{x\in A}$. This has a natural decomposition as a cubical complex $C$, with the vertices of cubes occuring at $\mathbb{Z}^A\subset \mathbb{R}^A$.
Let $C^{(1)}$, denote the 1-skeleton of $C$. Then: $$F_2''=\pi_1\left(C^{(1)}\right)$$ and killing off the conjugation action of $F_2''$ on itself we get:
$$F_2''/[F_2'',F_2'']=H_1\left(C^{(1)}\right)$$
As $$H_1\left(\mathbb{R}^A\right)=0,$$ we have $H_1\left(C^{(1)}\right)$ generated (as an Abelian group) by the boundaries of squares in $C^{(2)}$.
As $$H_2\left(\mathbb{R}^A\right)=0,$$ we know that the relations between these generators are generated by the boundaries of cubes in $C^{(3)}$.
Killing off the conjugation action of $F_2'/F_2''$ on $F_2''/[F_2'',F_2'']$ we get: $$F_2''/[F_2'',F_2']=H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$$ where ${C_\infty}^{\!\!A}\cong F_2'/F_2''$ acts on $H_1\left(C^{(1)}\right)$ by translating boundaries of squares in the natural way.
Thus $H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$ is generated by boundaries of squares with the origin and $v_x+v_y$ as opposite vertices, which we may index $\{s_{\{x,y\}}\}_{\{x,y\}\in A^{(2)}}$.
As the boundary of a 3 dimensional cube consists of pairs of parallel squares, with ${\it opposite}$ orientations, we have $$H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$$ freely generated by the $\{s_{\{x,y\}}\}_{\{x,y\}\in A^{(2)}}$ as an abelian group.
Finally we kill off the conjugation action of $A=F_2/F_2'$ on $F_2''/[F_2'',F_2']$. The conjugation action of $z\in A$ on $s_{\{x,y\}}$ is given by: $$zs_{\{x,y\}}=s_{\{x+z,y+z\}}$$
Thus $F_2''/[F_2'',F_2]$ is freely generated as an abelian group by $\{s_{\{0,x\}}\}_{x\in A}$, which is infinite. Any set of elements which normally generates $F_2''$ would generate this abelian group, so must be infinite.