I am having some trouble understanding what happens towards the bottom of this image. It is my understanding that $\mathbf{y}$ is a vector of length $1$; I can't quite see how this can be converted to the sum after the second equality. Furthermore (taking that as granted), why does the maximization condition as expressed in the last sentence hold true? Why can't $\mathbf{y}_j \geq 0$ for the other eigenvalues?
The second equation follows from $\mathbf{y} = \begin{pmatrix} y_1 & \dots & y_n \end{pmatrix}^T$, the definition of the norm $\newcommand{\norm}[1]{\left\lVert#1\right\rVert} \norm{\mathbf{y}} = \sqrt{y_1^2 + \dots + y_n^2}$, the statement $\norm{\mathbf{y}}=1$ being equivalent to $y_1^2 + \dots + y_n^2=1$ and $\mathbf{\Lambda} = \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}$.
Furthermore, I assume that you want to know why $\mathbf{y}_j > 0$ for a $j \not\in I$ can not be true (if we only assume $\mathbf{y}_j \geq 0$ as you wrote, it could still be $0$).
For the maximamization condition, observe first that $$ \max_{y_1^2+\dots+y_n^2=1} \sum_{i=1}^n \lambda_i y_i^2 \leq \max_{y_1^2+\dots+y_n^2=1} \sum_{i=1}^n \lambda_{\max}(\mathbf{A}) y_i^2 = \lambda_{\max}(\mathbf{A}) \sum_{i=1}^n y_i^2 = \lambda_{\max}(\mathbf{A}). $$
If we choose $\mathbf{y}$ as defined in the last sentence, we have $$ \sum_{i=1}^n \lambda_i y_i^2 = \sum_{i \in I} \lambda_{\max}(\mathbf{A}) y_i^2 = \lambda_{\max}(\mathbf{A}) \sum_{i \in I} y_i^2 = \lambda_{\max}(\mathbf{A}), $$ meaning that we have equality indeed.
However, if we have some $j \not\in I$ so that $y_j > 0$, then $\lambda_j < \lambda_{\max}(\mathbf{A})$ by definition of $I$. But then $$ \sum_{i=1}^n \lambda_i y_i^2 = \lambda_j y_j^2 + \sum_{i=1 \\ i \neq j}^n \lambda_i y_i^2 < \lambda_{\max}(\mathbf{A}) y_j^2 + \sum_{i=1 \\ i \neq j}^n \lambda_{\max}(\mathbf{A}) y_i^2 = \sum_{i=1}^n \lambda_{\max}(\mathbf{A}) y_i^2 = \lambda_{\max}(\mathbf{A}) $$ (the last equality is like the one we computed before), so we have strict inequality and we can not have a maximum.