Here we have
Riemann surface $X$, connected but not necessarily compact,
meromorphic, possibly constant but with no removable singularities, $f: X \to \mathbb C$, i.e. holomorphic $f: X \setminus \text{poles}(f)\to \mathbb C$, corresponding to
holomorphic $F: X \to \mathbb C_{\infty}$, where $F(x):=f(x), x \in \text{Domain}(f)$, $F^{-1}(\infty):=\text{poles}(f)$ and where $F$ is not identically $\infty$
From Rick Miranda - Algebraic Curves and Riemann surfaces
I believe I understand (though not so deeply) the statement of Lemma II.4.7, but I'm really confused about the proof.
Question: How does Lemma II.4.4 give us $mult_p F = ord_p(f-f(p))$, when $p$ is not a pole of $f$ (i.e. $f$ is holomorphic at $p$, since I guess $f$ has no removable singularities)? In particular, I want to know the relationship of the 'h' s.t. $mult_pF = 1 + ord_{z_o}(\frac{dh}{dz})$ and the 'f' that $F$ corresponds to
If $g$ is a holomorphic function on an open subset of $\mathbb{C}$ vanishing at $z_0$, then the order of $z_0$ as a zero of $g$ is the lowest integer $m$ such that $g(z) = (z - z_0)^m \times g_0(z)$ with $g_0$ is a holomorphic function with $g_0(z_0) \neq 0$. Differentiating the previous equality shows that if $z_0$ is a zero of $g$ \begin{align} \mathrm{ord}_{z_0}(g) = 1 + \mathrm{ord}_{z_0}(g') \end{align}
Note that if $C$ is a constant and if $f$ is a holomorphic function, then $(f+c)' =f' $.
Now, if $h$ is holomorphic near $z_0$, then $h-h(z_0)$ is holomorphic, vanishes at $z_0$, and \begin{align} \mathrm{ord}_{z_0}(h-h(z_0)) &= 1 + \mathrm{ord}_{z_0}\left((h-h(z_0))'\right) \\ &= 1 + \mathrm{ord}_{z_0} (h') \end{align}
Now suppose $f : X \to \mathbb{C}$ is meromorphic on the Riemann surface $X$. Let $p\in X$ be a point that is not a pole of $f$ and $\phi : U \subset X \to \phi(U)\subset \mathbb{C}$ is an holomorphic chart. Then $h = f\circ \phi^{-1} : \phi(U) \to \mathbb{C}$ is an holomorphic map on an open subset of $\mathbb{C}$, and if $z_0=\phi(p)$, \begin{align} \mathrm{mult}_pF &= \mathrm{ord}_p(f-f(p))\\ &= \mathrm{ord}_{z_0} (h-h(z_0)) & \text{because } \phi \text{ is invertible} \\ &= 1 + \mathrm{ord}_{z_0}\left((h-h(z_0))'\right) \\ &= 1 + \mathrm{ord}_{z_0}(h') \\ &= 1 + \mathrm{ord}_{p}(f') \end{align} because $h' = (f' \circ\phi^{-1})\times (\phi^{-1})'$, and as $\phi$ is a diffeomorphism, $h'$ has the same order of annihilation at $z_0$ than $f'$ at $p$.