Let $f_n(s) = \frac{1}{n^s}- \frac{1}{(n+1)^{s}}$ the derivative of this with respect to $s$ is as you computed $\frac{-\log n}{n^s}+ \frac{\log(n+1)}{(n+1)^{s}}$. This is a function of $s$, and you do not consider it on $[n,n+1]$, but rather $[1,s]$ for each $n$.
The MVT tells you that $\frac{f_{n}(s) - f_n(1)}{s -1} = f'(s_n)$ for some $s_n\in [1,s]$. Multiplying by $(s-1)$ and plugging in the explicit expressions for the functions this means $$(\frac{1}{n^s}- \frac{1}{(n+1)^{s}}) - (\frac{1}{n} -\frac{1}{n+1}) = (s-1)(\frac{-\log n}{n^s}+ n}{n^{s_n}}+ \frac{\log (n+1)}{(n+1)^{s}}).$$n+1)}{(n+1)^{s_n}}).$$ The left hand side appears in the original sum, and the result is obtained by instead plugging in the right hand side. 1 Since you edited the question but did not say that it is clear now, I assume you are hoping for some details in addition to what Ralph said. So: Let f_n(s) = \frac{1}{n^s}- \frac{1}{(n+1)^{s}} the derivative of this with respect to s is as you computed \frac{-\log n}{n^s}+ \frac{\log(n+1)}{(n+1)^{s}}. This is a function of s, and you do not consider it on [n,n+1], but rather [1,s] for each n. The MVT tells you that \frac{f_{n}(s) - f_n(1)}{s -1} = f'(s_n)  for some s_n\in [1,s]. Multiplying by (s-1) and plugging in the explicit expressions for the functions this means $$(\frac{1}{n^s}- \frac{1}{(n+1)^{s}}) - (\frac{1}{n} -\frac{1}{n+1}) = (s-1)(\frac{-\log n}{n^s}+ \frac{\log (n+1)}{(n+1)^{s}}). The left hand side appears in the original sum, and the result is obtained by instead plugging in the right hand side.