How does Ahlfors define derivative on a Riemann Surface? I'm reading a passage in Ahlfors (3rd Edition page 298) and he says the following:  

He has previously defined $G_0(f)$ to be the connected component of any germ generated by $f$. Then he wants to view $G_0(f)$ as the domain of a multivalued function $f$.
And $\pi$ is the projection from the sheaf to the complex numbers. So if $g$ is a germ centered at $z \in \mathbb{C}$, $\pi(g) = z$.  
What does he mean, saying $\theta: G_0(f) = G_0(g)$? Is it supposed to be $\theta: G_0(f) = G_0(g)\rightarrow G_0(f)$?  
In what way does this $g$ lead to the definition of a derivative of $f$? Why does $\theta$ have to have those 2 properties?
 A: Notation first: $\mathfrak{S}_0(\mathbf{f}) \neq \mathfrak{G}_0(\mathbf{f})$. It's $S_0$, not $G_0$. Not that it matters, of course.
What matters is that Ahlfors here isn't concerned with abstract Riemann surfaces, but with very special Riemann surfaces that come with a nice projection to $\mathbb{C}$.

What does he mean, saying $\theta: G_0(f) = G_0(g)$? Is it supposed to be $\theta: G_0(f) = G_0(g)\rightarrow G_0(f)$?

It is supposed to be $\theta \colon \mathfrak{S}_0(\mathbf{f}) \to \mathfrak{S}_0(\mathbf{g})$. I think it is nothing but a typo.

In what way does this $g$ lead to the definition of a derivative of $f$? Why does $\theta$ have to have those 2 properties?

It's not $\mathbf{g}$, it's $\theta$. Or, more precisely, the existence of such a $\theta$ when $\mathbf{g}$ is the global analytic function arising from the derivative of $\mathbf{f}$.
Abstractly, it is clear that a holomorphic map $\theta \colon X \to Y$ between two Riemann surfaces induces a homomorphism
$$\theta^\ast \colon \mathscr{O}(Y) \to \mathscr{O}(X)$$
by composition, $f \mapsto f\circ\theta$.
Now in the paragraph concerned, we have two Riemann surfaces $\mathfrak{S}_0(\mathbf{f})$ and $\mathfrak{S}_0(\mathbf{g})$ with the special property that the projection $\pi$ (restricted from the sheaf of germs of holomorphic functions to the respective surface) is a coordinate chart around every point of the surface. That means a map $\theta \colon \mathfrak{S}_0(\mathbf{f}) \to \mathfrak{S}_0(\mathbf{g})$ with the two properties $(1)$ and $(2)$ [note that since $\pi$ is a local homeomorphism, continuity of $\theta$ together with property $(1)$ implies that $\theta$ is a local homeomorphism] is a holomorphic map between the two Riemann surfaces.
We need the continuity of $\theta$ since a holomorphic map between Riemann surfaces is in particular continuous, and what we want is a holomorphic map between Riemann surfaces. We need that $\theta$ be fibre-preserving (that is, $\pi\bigl\lvert_{\mathfrak{S}_0(\mathbf{g})}\circ\,\theta = \pi\bigl\lvert_{\mathfrak{S}_0(\mathbf{f})}$) on the one hand to obtain the holomorphicity of $\theta$ (from the continuity, since $\pi$ is a local chart), to guarantee that $\theta$ is not constant (every constant map from $\mathfrak{S}_0(\mathbf{f})$ to any other Riemann surface is holomorphic, of course, but neither that map nor its induced pull-backs are very interesting), and so that each point and its image in the Riemann surfaces correspond to the same point in the plane. We get a relation between the branches of $\mathbf{f}$ and the branches of $\mathbf{g}$ only when the map $\theta$ preserves base-points.
Since $\mathbf{g}$ is a holomorphic function on $\mathfrak{S}_0(\mathbf{g})$, its pull-back $\theta^\ast(\mathbf{g}) = \mathbf{g}\circ\theta$ is a holomorphic function on the Riemann surface $\mathfrak{S}_0(\mathbf{f})$ by the above abstract consideration.
For the derivatives, one must then see that there is such a map $\theta \colon \mathfrak{S}_0(\mathbf{f}) \to \mathfrak{S}_0(\mathbf{f'})$. [The case of the higher derivatives follows by iteration/composition of the maps $\theta_n \colon \mathfrak{S}_0(\mathbf{f^{(n)}}) \to \mathfrak{S}_0(\mathbf{f^{(n+1)}})$.]
But, if the germ of $f$ in $z_0$ can be analytically continued along a path $\gamma$, then of course the germ of $f'$ in $z_0$ can be analytically continued along $\gamma$ too. And that furnishes the map $\theta$. If $f_\zeta = (f,\Omega) \in \mathfrak{S}_0(\mathbf{f})$ is a germ/function element obtained by analytic continuation of the germ $(f_0,\Omega_0)$ along a path $\gamma$, then $\theta(f_\zeta)$ is the germ $f'_\zeta$ obtained by analytic continuation of $(f_0',\Omega_0)$ along $\gamma$.
