In Elias Stein's Real Analysis, at the beginning of Chapter 4.1, it reads "We shall say that $f$ is finite-valued if $-\infty<f(x)<\infty$ for all $x$."

A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.

Since a valued function may have $\mathbb R \cup \{\infty\}$ as target, it's possible that finite function $f$ corresponds to cases where $\forall x \in \mathbb R \quad f(x) \neq \infty$ like $f(x)=x$ or $f(x)= \frac x {x^2+6}$ while $f(x)=\frac 1x$ , for example, is not finite according to this meaning, because $f(0)=\infty$ (thing that can be taken by defintion or convention)

A convex function $f$ is said to be proper if its epigraph is non-empty and contains no vertical lines, i.e., if $f(x)<+\infty$ for at least one $x$ and $f(x)>-\infty$ for every x. (Section 4, Chapter 1, Convex analysis, Rockafellar, 1997)

I am not familiar with the term finite in this context. One possible definition would be this.

A function $f: A \to B$ is finite if and only if $f(A) \subseteq B$ is finite.

However, I would not use this definition because the relation $f \subseteq A \times B$ is still an infinite set (if $A$ is infinite).

No, It means there are only finitely many $y$ such that $f(x)=y$, for example $f(x)=0,1,2,\dots,n$ where $n\in\mathbb{N}$ is finitely valued, although it is also bounded, but not all bounded functions are finitely valued for example $f(x)=x$ on $\mathbb{R}$ or $\mathbb{Q}$ takes uncountably many values within any $|x_i-x_j|<\epsilon$