You are correct in saying that these theorems are essentially the same.
The mean value theorem is a general form of the Roll's theorem where the slope of secant is not necessarily zero.
Both theorems state that at some point the slope of tangent is the same as slope of the secant connecting the points (a , f(a) )and (b, f(b)). ~ stated in the answer.
I had bold that line cause, I am confused of that line. He stated that slope of tangent is the same as secant connecting the points (both theorems state that). Did he mean both theorem says that slope of tangent is same of both theorem? As I know, according to Rolle's theorem slope is 0. But, according to mean value theorem slope is $\frac{f(b)-f(a)}{b-a}$
The bolded part is right as it is. You're right the slope of the secant line is $\frac{f(b)-f(a)}{b-a}$, and the MVT says exactly that there is some $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. i.e there is some point $c\in (a,b)$ where the slope of the tangent line at $c$ equals the slope of the secant joining $(a,f(a))$ and $(b,f(b))$.
Also, to answer the question in the title, both theorems are equivalent (they each imply the other) so neither theorem is more general than the other. It is easy to see MVT implies Rolle's theorem (because $f(b)-f(a)=0$), but Rolle's theorem also implies MVT, simply by "rotating the picture" appropriately (i.e you apply Rolle's theorem to "flattened function" defined as $h(x)=f(x)-\left[\frac{f(b)-f(a)}{b-a}\right](x-a)$).