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 $f_T(Z)$.
You can try having a look at this paper:
http://arxiv.org/abs/math.AT/0011121
It's the most functorial-minded paper on algebraic geometry I'ver ever seen. It's written by an algebraic topologist. He cares mostly about affine and formal schemes.
The definition you're looking for is in section 4 of the paper. The functorial point of view for a formal scheme is a small filtered colimit of schemes, the colimit taken in the functor category.
Best Answer
Check out the paper of Kontsevich-Rosenberg Noncommutative spaces and Noncommutative Grassmannian and related constructions. You will get what you want. i.e. the definition of properness and separatedness of presheaves(as functors, taken as "space") and morphism between presheaves(natural transformations).
Notice that these definitions are general treatment for algebraic geometry in functorial point of view,nothing to do with noncommutative.
Definition for separated morphism and separated presheaves
Let $X$ and $Y$ be presheaves of sets on a category $A$(in particular, $CRings^{op}$). We call a morphism $X\rightarrow Y$ separated if the canonical morphism
$X\rightarrow X\times _{Y}X$ is closed immersion We say a presheaf $X$ on $A$ is separated if the diagonal morphism: $X\rightarrow X\times X$ is closed immersion
Definition for strict monomorphism and closed immersion
For a morphism $f$: $Y\rightarrow X$ of a category $A$, denote by $\Lambda _{f}$ the class of all pairs of morphisms
$u_{1}$,$u_{2}$:$X\Rightarrow V$ equalizing $f$, then $f$ is called a strict monomorphism if any morphism $g$: $Z\rightarrow X$ such that $\Lambda_{f}\subseteq \Lambda_{g}$ has a unique decomposition: $g=f\cdot g'$
Now we have come to the definition of closed immersion: Let $F,G$ be presheaves of sets on $A$. A morphism $F\rightarrow G$ a closed immersion if it is representable by a strict monomorphism.
Example
Let $A$ be the category $CAff/k$ of commutative affine schemes over $Spec(k)$, then strict monomorphisms are exactly closed immersion(classcial sense)of affine schemes. Let $X,Y$ be arbitrary schemes identified with the correspondence sheaves of sets on the category $CAff/k$. Then a morphism $X\rightarrow Y$ is a closed immersion iff it is a closed immersion in classical sense(Hartshorne or EGA)
Definition for proper morphism just follows the classical definition: i.e. universal closed and separated. You can also find the definition of universal closed morphism in functorial flavor in the paper I mentioned.