The following is from Lectures on Riemann Surfaces by O. Forster:
7.8. Theorem. Suppose $X$ is a Riemann surface, $a\in X$ and $\varphi\in\mathcal{O}_a$ is a holomorphic function germ at the point $a$. Then there exists a maximal analytic continuation $(Y, p,f, b)$ of $\varphi$.
Pʀᴏᴏғ. Let $Y$ be the connected component of $|\mathcal{O}|$ containing $\varphi$. Let $p$ also denote the restriction of the mapping $p: |\mathcal{O}| \to X$ to $Y$. Then $p: Y \to X$ is a local homeomorphism. By Theorem (4.6) there is a complex structure on $Y$ so that it becomes a Riemann surface and the mapping $p: Y \to X$ is holomorphic. Now define a holomorphic function $f: Y \to \mathbb{C}$ as follows. By definition every $\eta\in Y$ is a function germ at the point $p(\eta)$. Set $f(\eta) := \eta(p(\eta))$. One easily sees that $f$ is holomorphic and $p_*(p_\eta(f)) = \eta$ for every $\eta\in y$. Thus if one lets $b := \varphi$, then $(Y, p,f, b)$ is an analytic continuation of $\varphi$.
Now we will show that $(Y, p,f, b)$ is a maximal analytic continuation of $\varphi$. Suppose $(Z, q, g, c)$ is another analytic continuation of $\varphi$. Define the map $F: Z \to Y$ as follows. Suppose $\zeta \in Z$ and $q(\zeta) =: x$. By Lemma (7.7) the function germ $q_*(p_\zeta(g))\in\mathcal{O}_x$ arises by analytic continuation along a curve from $a$ to $x$ from the function germ $\varphi$. By Lemma (7.2) $Y$ consists of all function germs which are obtained by the analytic continuation of $\varphi$ along curves. Hence there exists exactly one $\eta\in Y$ such that $q_*(p_\zeta(g))=\eta$. Let $F(\zeta) = \eta$. It is easy to check that $F: Z \to Y$ is a fiber-preserving holomorphic map such that $F(c)=b$ and $F^*(f)=g$. $\square$
I don't understand the second paragraph of the proof. Why is the map $F:Z\to Y$ well-defined? If we choose a different curve from $a$ to $x$, we might get a different $\eta$...
For the relevant definitions:
Suppose $X$ and $Y$ are Riemann surfaces and $\mathcal{O}_X$ and $\mathcal{O}_Y$ are the sheaves of holomorphic functions on them. Suppose $p: Y \to X$ is an unbranched holomorphic map. Since $p$ is locally biholomorphic, for each $y\in Y$ it induces an isomorphism $p^*: \mathcal{O}_{X, p(y)} \to \mathcal{O}_{Y, y}$. Let
$$ p_*: \mathcal{O}_{Y, y} \to \mathcal{O}_{X, p(y)} $$
be the inverse of $p^*$.
7.6. Definition. Suppose $X$ is a Riemann surface, $a\in X$ is a point and $\varphi\in\mathcal{O}_a$ is a function germ. A quadrupel $(Y, p,f, b)$ is called an analytic continuation of $\varphi$ if:
- $Y$ is a Riemann surface and $p:Y \to X$ is an unbranched holomorphic map.
- $f$ is a holomorphic function on $Y$.
- $b$ is a point of $Y$ such that $p(b) = a$ and $$ p_*(p_b(f)) = \varphi.$$
An analytic continuation $(Y, p,f, b)$ of $\varphi$ is said to be maximal if it has the following universal property. If $(Z, q, g, c)$ is any other analytic continuation of $\varphi$, then there exists a fiber-preserving holomorphic mapping $F: Z\to Y$ such that $F(c) = b$ and $F^*(f) = g$.
Please note that it does not read $p_\zeta(g)$ but $\rho_\zeta(g)$ (rho), the germ of $g$ at the point $\zeta$.
The path chosen in $Z$ from $c$ to $\zeta$ is not decisive. Important is the germ $\rho_\zeta(g) \in \mathcal{O}_{Z,\zeta}$. Apparently it is independent from the path. It projects to the germ $$\eta:=q_*(\rho_\zeta(g)) \in \mathcal{O}_{X,x}$$ The latter is obtained as analytic continuation of $\phi$ along a curve - here you need the existence of at least one curve. Then by definition $\eta$ is a point of $Y$ and one defines $$F(\zeta):=\eta.$$