Assume that $f$ is a holomorphic function at $p \in X$ ($X$ Riemann surface) and that $mult_p(F)=1$. Is it true that $f$ is a complex chart? And in general every holomorphic map from an open set of a Riemann surface $X$ having constant multiplicity one is a complex chart?
In order to answer to the first question, I use the normal form of f. So, there exists a neighborhood of $p$ for which $f$ assumes the form $z \mapsto z$. It is clearly continuous and invertible, but what can I say of the inverse map?
Concerning the second question, I think the answer is no. Can you provide a counterexample, please?
Let $U$ be an open subset of a Riemann surface $X$ and $f: U\to {\mathbb C}$ an injective (aka one-to-one) holomorphic function. Then $f$ is a homeomorphism to its image. This is immediate from the property that nonconstant holomorphic functions (with connected domain) are open mappings.
Actually, one has more, namely, that $f: U\to f(U)\subset {\mathbb C}$ is a biholomorphic mapping, i.e. its inverse is also holomorphic. There are several ways to see this. One is to use the normal form of holomorphic functions (which, I think, you are familiar with):
Lemma. Let $h: D\to {\mathbb C}$ be a holomorphic function defined on an open subset $D$ in ${\mathbb C}$. Then for every $p\in D$ there exists a pair of biholomorphic (to their images) functions $\phi, \psi$ defined on neighborhoods $V$ and $W$ of $p$ and $f(p)$ respectively, such that $$ \psi\circ h \circ \phi^{-1} $$
has the form $z\mapsto z^n$, where $n\in {\mathbb N}\cup \{0\}$.
This lemma (assuming $h(p)=0$) is proven by writing $h$ (near $p$) in the form $h(z)= (p-z)^n g(z)$, where $g$ is a nonvanishing holomorphic function near $p$. Then consider a branch of the $n$-th root of $h(z)$: $$ h^{1/n}(z)= (p-z) (g(z))^{1/n} $$ and observing that this holomorphic function has nonzero derivative at $p$.
Applying this lemma in your situation and taking into account that your function $f$ is injective, we conclude that the number $n$ in this lemma equals to $1$ at every $p\in U$. Hence, $f$ restricts to a biholomorphic function on a neighborhood of every point $p\in U$. (Since your $f$, locally, is a composition of three biholomorphic functions.) Hence, $f^{-1}: f(U)\to U$ is holomorphic near each point $q\in f(U)$, hence, $f^{-1}$ is holomorphic.