You can prove this by repeatedly applying the following lemma:

**Lemma**:
Let $G$ be a group and let $A,B \lhd G$ be finitely generated normal subgroups.  Assume that $A$ and
$B$ are commensurable.  Then $[G,A]$ and $[G,B]$ are commensurable and
$$\left(A / [G,A]\right) \otimes \mathbb{Q} \cong \left(B / [G,B]\right) \otimes \mathbb{Q}.$$

At the kth stage of the argument, you've proven that $\gamma_k(G)$ is a finite-index subgroup of $\gamma^{tf}_k(G)$.  Applying the lemma, we deduce that
the ranks of $\gamma_k(G)/\gamma_{k+1}(G)$ and $\gamma^{tf}(G)/[G,\gamma^{tf}_{k}(G)]$ are the same, and also that $\gamma_{k+1}(G)$ is a finite-index subgroup of $[G,\gamma^{tf}_{k+1}(G)]$.  By construction, the ranks of $\gamma^{tf}(G)/[G,\gamma^{tf}_{k}(G)]$ and $\gamma^{tf}(G)/\gamma^{tf}_{k+1}(G)$ are the same (giving one conclusion!), and also that $\gamma^{tf}_{k+1}(G)$ is commensurable with $[G,\gamma^{tf}_{k}(G)]$ (letting us continue the induction!).

**Proof of lemma**:
We can assume without loss of generality that $A$ is a finite-index subgroup of $B$.  The purported isomorphism can be rewritten as
$$H_1(A;\mathbb{Q})_G \cong H_1(B;\mathbb{Q})_G,$$
where the subscripts indicate that we are taking the $G$-coinvariants.  Since $B/A$ is a finite group, the Hochschild-Serre spectral sequence of the extension
$$1 \rightarrow A \rightarrow B \rightarrow B/A \rightarrow 1$$
degenerates to show that
$$H_k(B;\mathbb{Q}) \cong H_0(B/A;H_k(A;\mathbb{Q})) = H_k(A;\mathbb{Q})_B \quad \text{for all $k \geq 0$},$$
and taking the $G$-convariants of this gives that
$$H_k(A;\mathbb{Q})_G = \left(H_k(A;\mathbb{Q})_B\right)_G \cong H_k(B;\mathbb{Q})_G \quad \text{for all $k \geq 0$}.$$
The special case $k=1$ of this is what we were supposed to prove.  

To see that $[G,A]$ and $[G,B]$ are
commensurable, we will need the **Hirsch index** of a general group $\Gamma$, which is the supremum of the $n$ such that there exists a chain
$$\Gamma_0 < \Gamma_1 < \cdots < \Gamma_n$$
of subgroups of $\Gamma$ such that $\Gamma_i$ is an infinite-index subgroup of $\Gamma_{i+1}$ for all $0 \leq i < n$ (this is usually only defined for polycyclic groups, but it is not hard to see that this reduces to the usual definition in that case; see <a href="https://math.stackexchange.com/a/1000546">this</a> answer for more details).

Let $r$ be the common dimension of $\left(A / [G,A]\right) \otimes \mathbb{Q}$ and $\left(B / [G,B]\right) \otimes \mathbb{Q}$.
Since $A$ is a finite-index normal subgroup of $B$, we can apply the additivity result proved <a href="https://math.stackexchange.com/a/1000546">here</a> to the short exact sequence
$$1 \longrightarrow A/[G,A] \longrightarrow B/[G,A] \longrightarrow B/A \longrightarrow 1$$
to deduce that
$$h(B/[G,A]) = h(A/[G,A]) + h(B/A) = r + 0 = r.$$
Applying this additivity now to the short exact sequence
$$1 \longrightarrow [G,B]/[G,A] \longrightarrow B/[G,A] \longrightarrow B/[G,B] \longrightarrow 1,$$
we see that
$$r = h(B/[G,A]) = h([G,B]/[G,A]) + h(B/[G,B]) = h([G,B]/[G,A]) + r.$$
It follows that $h([G,B]/[G,A])=0$, so $[G,A]$ is a finite-index subgroup of $[G,B]$, as desired.