If we start with the Taylor expansion (I'll change variables here, too many subscripts confuse me):
$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}f^{\prime\prime}(x)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$
and the derivative of this w.r.t. $h$
$f^{\prime}(x+h)=f^{\prime}(x)+h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$
and the version of this with $h$ replaced by negative $h$:
$f^{\prime}(x-h)=f^{\prime}(x)-h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$
subtracting the third expression from the second expression gives
$f^{\prime}(x+h)-f^{\prime}(x-h)=2h f^{\prime\prime}(x)+O(h^3)$
and we see that the even powers drop out of this error expansion.
If we solve for $f^{\prime\prime}(x)$ like so:
$f^{\prime\prime}(x)=\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)$
and substitute in the first expression,
$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$
we can take the $O(h^2)$ within the parentheses out as an $O(h^4)$ term:
$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$
the leading term after the replaced portion is $O(h^3)$, thus simplifying to
$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+O(h^3)$
and we see that the formula has $O(h^3)$ error: cutting $h$ in half decreases the error by a factor of $2^3=8$.