Qcqs lemma in Ravi Vakil's notes

You might find it helpful to have a look at Hartshorne's $\textit{Algebraic Geometry}$, Chapter II, Exercise $2.16$. In brief:

The map $\Gamma(X,\mathscr{O}_X)_s \to \Gamma(X_s,\mathscr{O}_X)$ being injective means that given a section $a\in \Gamma(X,\mathscr{O}_X)$ whose restriction to $X_s$ is zero, there exists some $n$ such that $s^n \cdot a=0$.

For proving surjectivity, note that the intersection of two open affines in $X$ is quasi-compact since $X$ is quasi-separated. Use this to prove that multiplying a section of $\mathscr{O}_X$ over $X_s$ by a suitable power of $s$ you obtain an element which is the restriction of a global section.

If you need some further details, let me know.

$\textbf{Edit}$ For proving the part about surjectivity, let $X=\bigcup_{i=1}^n \operatorname{Spec}(A_i)$ be a finite open affine covering of $X$ and let $a\in \Gamma(X_s,\mathscr{O}_X)$. Since the statement we want to prove is true for affine schemes (the restriction map to $X_s$ is simply the localization map in this case), there exists some $m$ (independent of $i=1,\ldots,n$) and elements $a_i\in A_i$ such that $s^m \vert_{X_s \cap \operatorname{Spec}(A_i)} \cdot a\vert_{X_s \cap \operatorname{Spec}(A_i)} = a_i\vert_{X_s\cap \operatorname{Spec}(A_i)}$. In particular, $a_i\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)} - a_j\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)}\in \Gamma(\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j),\mathscr{O}_X)$ restricts to $0$ over $X_s\cap \operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)$. Since the intersection of two open affines of $X$ is quasi-compact, we conclude from the injectivity part that there exists some $M$ (independent of $i,j$) such that $$ s\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)}^M \cdot a_i\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)} = s\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)}^M \cdot a_j\vert_{\operatorname{Spec}(A_i)\cap \operatorname{Spec}(A_j)} $$ for all $i,j=1,\ldots,n$. In other words, the sections $s\vert_{\operatorname{Spec}(A_i)} \cdot a_i$ glue to a section $b$ and since $$ b\vert_{X_s\cap \operatorname{Spec}(A_i)} = (s^M\vert_{\operatorname{Spec}(A_i)} \cdot a_i)\vert_{X_s\cap \operatorname{Spec}(A_i)}= s^{M+m}\vert_{X_s\cap \operatorname{Spec}(A_i)} \cdot a\vert_{X_s\cap \operatorname{Spec}(A_i)} $$ for $i=1,\ldots,n$, we conclude that $b\vert_{X_s}=s^{M+m}\vert_{X_s} \cdot a$.