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 $f(U_n) \subset V_m$.
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.
