Yes, a universally closed morphism is quasi-compact. (I haven't yet checked whether the same approach answers question 2.)
Proof: Without loss of generality, we may assume that $S=\operatorname{Spec} A$ for some ring $A$, and that $f$ is surjective. Suppose that $f$ is not quasi-compact. We need to show that $f$ is not universally closed.
Write $X = \bigcup_{i \in I} X_i$ where the $X_i$ are affine open subschemes of $X$. Let $T=\operatorname{Spec} A[\{t_i:i \in I\}]$, where the $t_i$ are distinct indeterminates. Let $T_i=D(t_i) \subseteq T$. Let $Z$ be the closed set $(X \times_S T) - \bigcup_{i \in I} (X_i \times_S T_i)$. It suffices to prove that the image $f_T(Z)$ of $Z$ under $f_T \colon X \times_S T \to T$ is not closed.
There exists a point $\mathfrak{p} \in \operatorname{Spec} A$ such that there is no neighborhood $U$ of $\mathfrak{p}$ in $S$ such that $X_U$ is quasi-compact, since otherwise we could cover $S$ with finitely many such $U$ and prove that $X$ itself was quasi-compact. Fix such $\mathfrak{p}$, and let $k$ be its residue field.
First we check that $f_T(Z_k) \ne T_k$. Let $\tau \in T(k)$ be the point such that $t_i(\tau)=1$ for all $i$. Then $\tau \in T_i$ for all $i$, and the fiber of $Z_k \to T_k$ above $\tau$ is isomorphic to $(X - \bigcup_{i \in I} X_i)_k$, which is empty. Thus $\tau \in T_k - f_T(Z_k)$.
If $f_T(Z)$ were closed in $T$, there would exist a polynomial $g \in A[\{t_i:i \in I\}]$ vanishing on $f_T(Z)$ but not at $\tau$. Since $g(\tau) \ne 0$, some coefficient of $g$ would have nonzero image in $k$, and hence be invertible on some neighborhood $U$ of $\mathfrak{p}$. Let $J$ be the finite set of $j \in I$ such that $t_j$ appears in $g$. Since $X_U$ is not quasi-compact, we may choose a point $x \in X - \bigcup_{j \in J} X_j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $P \in T$ lying above $u$ such that $g(P) \ne 0$ and $t_i(P)=0$ for all $i \notin J$. Then $P \notin T_i$ for each $i \notin J$. A point $z$ of $X \times_S T$ mapping to $x \in X$ and to $P \in T$ then belongs to $Z$. But $g(f_T(z))=g(P) \ne 0$, so this contradicts the fact that $g$ vanishes on $f_T(Z)$.
Begin with $Y$ equal to the affine plane, $\text{Spec}\ R$, for $R=k[s,t]$. Let $\overline{f}:\overline{X}\to Y$ be the blowing up of $Y$ at the ideal $\mathfrak{m} = \langle s,t \rangle$. Define $I\subset \mathfrak{m}$ to be the ideal $\langle s,t^2 \rangle$. Define $\mathcal{G}$ to be $\widetilde{R/\mathfrak{m}}$, define $\mathcal{F}$ to be $\widetilde{R/I}$, and define $p:\mathcal{F}\to \mathcal{G}$ to be the natural surjection. The pullback $\overline{f}^*\mathcal{G}$ is the structure sheaf of the exceptional divisor $E$. The pullback $$\overline{f}^*p:\overline{f}^*\mathcal{F}\to \overline{f}^*\mathcal{G},$$ is an isomorphism except at a single point $q$ on $E$.
Define $X$ to be the open complement of $\{q\}$ in $\overline{X}$. Define $f:X\to Y$ to be the restriction of $\overline{f}$ to $X$. Since we remove a single closed point from the positive dimensional fiber $E$ of $\overline{f}$, the morphism $f$ is still surjective. On $X$, the natural homomorphism $$f^*p:f^*\mathcal{F}\to f^*\mathcal{G},$$ is an isomorphism. Therefore, the pullback map $H^0(Y,\mathcal{F}) \to H^0(X,f^*\mathcal{F})$ factors through the map $$H^0(p):H^0(Y,\mathcal{F})\to H^0(Y,\mathcal{G}).$$ This map has a one-dimensional kernel.
Best Answer
A morphism $f: X \to Y$ of schemes is universally injective (radicial) iff it is injective and for all $y = f(x)$ the extension $k(y) \subseteq k(x)$ is a purely inseparable algebraic extension. (EGA I, (3.5.8))
So for schemes of finite type over $\mathbf{C}$ , an injective morphism is universally injective iff for all $y = f(x)$ with $x$ non-closed the extension $k(y) \subseteq k(x)$ is a purely inseparable algebraic extension, i.e. $k(y) = k(x)$. Perhaps this is automatically the case if it holds for all closed points and everything is of finite type, so we have Jacobson schemes.