Are holomorphic functions between Riemann surfaces continuous? From differential topology, we know that every smooth map between smooth manifolds is continuous. But for the definition of holomorphic functions between Riemann surfaces as follows

I do not think that the map is guaranteed to be continuous since it does not have the critical condition that $f(U_n) \subset V_m$.
 A: This is a subtle point that many authors overlook.
It depends on precisely what you mean by saying that a function $f$ defined on an arbitrary subset $A\subset \mathbb C$ is analytic. One common definition is that the domain $A$ must be an open subset of $\mathbb C$ and $f$ has a complex derivative everywhere in $A$ (or is given locally by convergent power series). Using this definition, saying that $g_m\circ f \circ f_n^{-1}\colon f_n (U_n \cap f^{-1}(V_m))\to g_m(f(U_n)\cap V_m)$ is analytic includes the tacit requirement that $ f_n (U_n \cap f^{-1}(V_m))$ is open in $\mathbb C$ for each $m$ and $n$. Using this interpretation, if $f\colon S_1\to S_2$ is analytic according to the quoted definition, then it is automatically continuous.
To see why, let $p\in S_1$ be arbitrary. There exist charts $U_n$ containing $p$ and $V_m$ containing $f(p)$, and then
$$
f_n (U_n \cap f^{-1}(V_m)) \text{ is open in $\mathbb C$}
\implies 
U_n \cap f^{-1}(V_m)\text{ is open in $S_1$}
$$
(because $f_n$ is a homeomorphism onto its image). Restricted to the open set $U_n \cap f^{-1}(V_m)$, $f$ is given by the formula
$$
f= g_m^{-1} \circ \big( g_m \circ f \circ f_n^{-1}\big) \circ f_n^{-1},
$$
which is a composition of continuous maps. Thus $f$ is continuous in a neighborhood of each point, and hence continuous on $S_1$.
If, on the other hand, your interpretation of "analytic on a subset of $\mathbb C$" does not necessarily entail being defined on an open subset, then an analytic map by this definition might not be smooth. For example, one might define a function $f$ to be analytic on a subset $A\subset\mathbb C$ if each point of $A$ has a neighborhood $U$ in $\mathbb C$ such that $f$ is equal to a convergent power series on $U\cap A$. 
Using this interpretation, a function $f\colon S_1\to S_2$ that is analytic by the quoted definition need not be continuous. Here's a counterexample. Let $S_1 = S_2  = \mathbb C$, with the analytic structure given by, say, the countable collection of disks with rational centers and radius $1/2$, and with $f_n$ and $g_m$ the respective identity maps. Define $f\colon S_1\to S_2$ by 
$$
f(z) = \begin{cases}
1, & \operatorname{Re}z\ge 0,\\
0, & \operatorname{Re}z <0.
\end{cases}
$$
Then for each $m$ and $n$ for which $U_n \cap f^{-1}(V_m)\ne\varnothing$, the composite map $g_m\circ f \circ f_n^{-1}$ is either identically $0$ or identically $1$ (but not always defined on an open set), so it is analytic by our broader definition. But $f$ is not continuous.
