You need some extra assumptions:$f$ not constant on any component of $X$, $Y$ connected, ($X$ compact, or some behavior of $f$ at infinity if $X$ not compact). The degree of the map $f\colon X \to Y$ is a positive integer $m$ so that:
for every $y$ in $Y$ the equation $f(x)=y$ has exactly $m$ solutions $x \in X$ ( counting multiplicities). The fact that this number is constant is a theorem.
There will be just finitely many points $y$ in $Y$ for which the multiplicities of some roots are $>1$.
A basic example is the map $z \mapsto z^m$ from $\mathbb{C}$ to $\mathbb{C}$ or even better, from $\hat{ \mathbb{C}}$ to $\hat{ \mathbb{C}}$. Another important example are polynomial maps of degree $m\ $
$$z \mapsto a_m z^m + \ldots + a_0$$
You forgot to say $F$ is non-constant. Then again, I guess $Ram(F)$ is not defined for $F$ non-constant.
In general for any map $F: X \to Y$ of any topological spaces $X$ and $Y$ with $X$ compact and $Y$ Fréchet/T1 and for any closed discrete subspace $A$ of $X$, we have $F(A)$ discrete.
Proof: Closed discrete subspaces $A$ of compact is finite $\implies$ $A$ is finite $\implies$ $F(A)$ is finite $\implies$ $F(A)$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED
Apply this to the case of $A=Ram(F)$ when $F$ is a non-constant holomorphic map between connected Riemann surfaces with $X$ compact (and thus $F$ is surjective, open, closed and proper and $Y$ is compact) to get $F(A)=Branch(F)$ is discrete.
In particular, this means we do not use that $F$ is proper, closed, open, surjective, non-constant or holomorphic or that $X$ is connected or that $Y$ is connected. We can relax this to $X$ compact (and not necessarily Riemann surface) and $Y$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).
Best Answer
Yes, it's just the obvious thing: $f$ is holomorphic at $x$ if whenever $\varphi$ is a complex chart of $X$ defined at $x$ and $\psi$ is a complex chart of $Y$ defined at $f(x)$, $\psi\circ f\circ\varphi^{-1}$ is holomorphic at $\varphi(x)$.