This follows from Zariski's main theorem if the characteristic is zero and it is false in positive characteristics: consider the the morphism $\mathbb{A}^1 \to \mathbb{A}^1$ given by $x \mapsto x^p$ where $p$ is the characteristic. The statement would also be true in char p if you assume that the general fibre is reduced.
(Note that it suffices to assume that X is integral and Y is normal. $f$ should of course also be surjective.)
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 $f_T(Z)$.
Best Answer
It might be helpful to split your question into two parts:
When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all $s \in S$, $X \to S$ satisfies this property?
When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all closed $s \in S$, all fibers satisfy this property.
The answer to the first question is almost never positive if you do not make additional assumptions. The two main cases, where I know of a positive answer, are:
Ia. The first one is Francesco's answer: If $X \to S$ is flat (EDIT: and locally of finite presentation) and all fibers are smooth (or etale), then $X \to S$ has the same property. There are variants for other types of singularities.
Ib. If $X \to S$ is proper and all fibers are finite, then $X \to S$ is finite (Grothendieck's version of Zariski's main theorem).
There are variants, as if $f\colon X \to S$ is proper, flat and of finite presentation and $X_s \to {\rm Spec} \kappa(s)$ is an isomorphism for one $s \in S$, then there exists an open neighborhood $U$ of $s$ such that $f^{-1}(U) \to U$ is an isomorphism.
The answer to the second question is much more often positive. There is the following general principle: Assume that $X$ is of finite presentation over $S$. Let $C$ be the set of points $s \in S$ where $X_s \to {\rm Spec} \kappa(s)$ has a certain "decent" property $P$. Usually "decent" should at least imply that you have a property of schemes over a field that is stable under extension of the base field; thus "irreducible" is not a decent property, but "geometrically irreducible" is.
Then $C$ is constructible. $C$ is open if $X$ is in addition proper and flat over $S$.
Of course this is not true for all properties (e.g., for the property of being an affine morphism) but it does hold for many properties. Many statements of this kind are proved in EGAIV, mainly §9 and §12. There is also a rather long list in the appendix of my book with Görtz.
Therefore to answer the second question it suffices to answer the following question.
In general the answer is no because there are schemes that do not contain any closed point. If $C$ is constructible, then the answer is positive if $S$ is of finite type over a field or, more generally, if $S$ is Jacobson (e.g. if $S$ is of finite type over a Dedekind ring with infinitely many prime ideals - as the ring of integers). In this case $C \mapsto C \cap S_0$ yields a bijection between constructible subsets of $S$ and constructible subsets of $S_0$.
If $C$ is open, then to get a positive answer it suffices to have a scheme where every point has a closed point in its closure. This is true for quasi-compact schemes (easy to see using that this is certainly true for affine schemes) or locally noetherian schemes (not that easy to see).