I can see why $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ holds: multiplying both sides of the inequality by $(k+1)$ allows me to simplify to the form $\frac{1}{k+1} \leq \frac{1}{k}$, but it's unclear to me why $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ is sufficient to show that $\sum_{i=1}^ki + \frac{1}{(k+1)^2} \leq \ 2-\frac{1}{k}+\frac{1}{(k+1^2)}$.
The thing is, we do not want to show that $$\sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k}+\frac{1}{(k+1)^2}.$$
Instead, the thing we must show in the inductive case is $$ \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1}. \tag1 $$
Prior to the point where you had your doubts, the proof had already shown that $$ \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2 + \left( -\frac{1}{k}+\frac{1}{(k+1)^2}\right). \tag2 $$ Can you see how Inequality $(2)$, plus the fact that $-\dfrac{1}{k}+\dfrac{1}{(k+1)^2} \leq -\dfrac{1}{k+1}$, is enough to prove Inequality $(1)$?