Let $$T_{f}(a,x) = \sup \sum_{j=1}^{n}|f(t_{j}) - f(t_{j-1})|$$ be the total variation of $f$ on $[a,x]$. I want show that:
$T_{f}(a,x)$ is absolutely continuous in $[a,b]$ whenever $f$ is absolutely continuous $[a,b]$.
My attempt. We need to show that for each $\epsilon > 0$, there is a $\delta>0$ such that:
$$\sum_{1}^{m}|T_{f}(a,b_{k}) - T_{f}(a,a_{k})| < \epsilon$$ whenever $$\sum_{1}^{m}|b_{k} - a_{k}|<\delta.$$
Note that $$\sum_{1}^{m}|T_{f}(a,b_{k}) - T_{f}(a,a_{k})| = \sum_{1}^{m}T_{f}(a,b_{k}) - T_{f}(a,a_{k}) = \sum_{1}^{m}T_{f}(a_{k},b_{k}).$$
Choose $\delta$ that satisfies the absolutely continuity of $f$ on $[a,b]$ and $\displaystyle \tilde{\epsilon} < \min\left\{\frac{\epsilon}{m+1},\frac{\epsilon}{2}\right\}$. So,
$$\sum_{1}^{n}|t_{j}-t_{j-1}|<\delta$$ implies $$\sum_{1}^{n}|f(t_{j}) - f(t_{j-1})|<\tilde{\epsilon}.$$ for any partition of $[a_{k},b_{k}]$. Therefore,
$$T_{f}(a_{k},b_{k}) = \sup \sum_{1}^{n}|f(t_{j}) - f(t_{j-1})|<\frac{\epsilon}{m}.$$
Thus,
$$\sum_{1}^{m}T_{f}(a_{k},b_{k}) <\epsilon.$$
That makes sense?
Your proof is fine. A slick way to do this is to note that since $f$ is absolutely continuous, we have
$1).\ T_f(a,x)=\int^x_a|f'(t)|dt,$
and so $T_f$ is absolutely continuous (use absolute continuity of the Lebesgue integral).
$1).$ follows from the following facts:
$2).\ T_f$ is nondecreasing, so $\int^x_aT_f'(a,x)\le T_f(a,x).$
$3).\ |f(x) - f(y)| \le |T_f(a,x) - T_f(a,y)|$, so $|f'|\le |T'_f|=T'_f.$
$2).$ and $3).$ combine to give
$4).\ \int|f'|dt\le T_f.$
On the other hand, $f$ is absolutely continuous, so
$5).\ \sum^n_{j=1} |f(t_{j+1}) - f(t_j)| \le |\sum^n_{j=1}\int^{t_j}_{t_{j-1}} f'(t)dt|=|\int^{b}_a f'(t)dt|\le \int^{b}_a |f'(t)|dt$.
Suping over the LHS of $5).$, we get
$6).\ T_f(a,x)\le \int |f'(t)|dt$.
$4).$ and $6).$ together give $1).$