The ordinary proof that 0 derivative implies constant rests on the mean value theorem. Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives. A quick google search came up with the paper
http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf
Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with
$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$
Clearly then if $f^s = 0$ we must have that $f$ is constant.
EDIT: For your original question, it is not true that BV + continuous + existence of symmetric derivative implies AC. You can imagine a function whose graph looks like follows:
in the interval [0,1/2], it is a wedge shape $\Lambda$ which isn't too tall or too steep.
in the interval [1/2,3/4], it is a wedge which is shorter and steeper.
in [3/4,7/8], it is even shorter and steeper.
etc.
By controlling the rates at which the heights and steepnesses change, it is possible to
(1) Get BV (make the heights decrease rapidly).
(2) Get continuity and symmetric differentiability at 1 (make the heights decrease rapidly). Continuity and symmetric differentiability at all points but $x=1$ is obvious.
(3) Cause AC to fail (let the steepnesses increase without bound).
In fact it is easy to see from the quasi-mean value theorem that if $f$ is continuous and $f^s$ exists everywhere then $f$ is AC if and only if $f^s$ is bounded. This is likely as good a necessary and sufficient condition as you can come up with, as we have seen that BV does not force $f^s$ to be bounded.