Neither statement involves a one-sided limit. In both statements $h$, whether as $h$ or as $-h$, can be either positive or negative.
Both statements are shorthand. The first says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a+h)-f(a)}h-f'(a)\right|<\epsilon\tag{1}$$ whenever $0<|h|<\delta$. The second says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a-h)-f(a)}{-h}-f'(a)\right|<\epsilon\tag{2}$$ whenever $0<|-h|<\delta$. Now $|-h|=|h|$, so in both cases you're simply saying that a certain inequality involving $f'(a)$ is true whenever $-\delta<h<0$ or $0<h<\delta$. There is no implication that $h$ is positive or that $-h$ is negative. The question is whether saying that $(1)$ is true whenever $-\delta<h<0$ or $0<h<\delta$ is the same thing as saying that $(2)$ is true whenever $-\delta<h<0$ or $0<h<\delta$.
The answer is yes. Let $$q_1(h)=\frac{f(a+h)-f(a)}h$$ and $$q_2(h)=\frac{f(a-h)-f(a)}{-h}\;.$$
A little algebra shows that $q_1(-h)=q_2(h)$ for all non-zero $h$. As $h$ takes on all non-zero values between $-\delta$ and $\delta$, so does $-h$; it just does so in reverse order, so to speak. Thus, saying (as in $(1)$) that $|q_1(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$ is exactly the same as saying that $|q_1(-h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$, which is the same as saying (as in $(2)$) that $|q_2(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$.
