Understanding notation for holomorphic quadratic differentials I've found the notation in books to be a bit confusing (perhaps my background is inadequate).
For example, in Farb and Margalit's book A Primer on Mapping Class Groups, they give the following definition:
Definition
"Let $\{ z_\alpha: U_\alpha \to \mathbb{C} \}$ be an atlas for $X$ [Here $X$ is a Riemann surface]. A holomorphic quadratic differential $q$ on $X$ is specified by a collection of expressions $\{ \phi_\alpha (z_\alpha) dz_\alpha^2 \}$ with the following properties:
1. Each $\phi_\alpha : z_\alpha(U_\alpha) \to \mathbb{C}$ is a holomorphic function with a finite number of zeros.
2. For any two coordinate charts $z_\alpha$ and $z_\beta$, we have
$$ \phi_\beta(z_\beta)( \frac{dz_\beta}{dz_\alpha})^2 = \phi_\alpha(z_\alpha) $$
"
Here $\frac{dz_\beta}{dz_\alpha}$ is the derivative of the change of coordinates  $z_\beta \circ z_\alpha^{-1}$.
So, my problem comes when I try to check that this definition is independent of the coordinate system that you choose. 
Say that you have $z_\alpha: U_\alpha \to \mathbb{C}, z_\beta: U_\beta \to \mathbb{C}$, $w \in U_\alpha \cap U_{\beta}$ and a vector $v \in T_wX$. Then, I would like to use property number 2 to check that $q(v)$ is well defined, i.e. that
$$ \phi_\alpha \circ z_\alpha(w)  dz_\alpha^2(v) = \phi_\beta \circ z_\beta(w)  dz_\beta^2(v) $$
I'm confused about what exactly $dz_\alpha(v)$ and $dz_\beta(v)$ are. Is it the case that $dz_\alpha = z_\alpha^*(dz)$? 
I thought that could be it, but then I get $dz_\alpha(v) = dz(d_w z_\alpha(v)) $, and I don't really get anywhere. I would like to do this in a rigorous way, not just formally manipulating $dz_\alpha$ and $dz_\beta$ ("multiplying" by $dz_\alpha^2$ in property number 2 would give me what I want, but I would like to understand what I'm doing).
 A: Since your confusion is on the definition of $dz_\alpha$ I will recall how we define a one-form on a Riemann surface.
Each $z_\alpha$ is a coordinate chosen for the open subset $V_\alpha = z_\alpha(U_\alpha) \subset \mathbb{C}$. Note that the map $z_\alpha\colon U_\alpha \rightarrow V_\alpha$ is a priori just a homeomorphism, it does not make sense (a priori) to talk about its derivative. We just have that the maps
$$
z_\beta\circ z_\alpha^{-1}\colon V_\alpha\cap V_\beta \rightarrow V_\alpha\cap V_\beta
$$
are biholomorphisms.
Now recall that if we have to coordinates $z,w$ on a open subset of $\mathbb{C}$ we may write a one form $\omega$ in the two coordinates and their expressions must agree
$$
f(z)dz = \omega = g(w)dw
$$
This means that if $w=T(z)$ for some biholomorphism $T$ we may have $T^\ast g(w)dw = f(z)dz$ i.e. $$g(T(z))T'(z) = f(z). \tag{1}$$
To define a one-form on a Riemann surface is to give a one form $\omega_\alpha$ on $V_\alpha$, for each $\alpha$  and impose that these forms agree on $V_\alpha \cap V_\beta \neq \emptyset$. Now if $\omega_\alpha = \phi_\alpha dz_\alpha$ we may use the previous remark to see that
$$
 \phi_\alpha dz_\alpha = (z_\beta\circ z_\alpha^{-1})^\ast \phi_\beta dz_\beta
$$
hence 
$$\phi_\beta \circ (z_\beta\circ z_\alpha^{-1}) \left( \frac{d(z_\beta\circ z_\alpha^{-1})}{dz_\alpha}\right) = \phi_\alpha.$$
Moreover we can compose with $z_\alpha$ to get 
$$\phi_\beta \circ z_\beta \left( \frac{d(z_\beta\circ z_\alpha^{-1})}{dz_\alpha}\right)\circ z_\alpha = \phi_\alpha\circ z_\alpha.$$
(It is just the equation $(1)$ with $z_\alpha = z$, $z_\beta = w$ and $z_\beta\circ z_\alpha^{-1} = T$).
Now you can define quadratic differentials forms in the same way.
After all we can write heuristically $dz_\alpha = z_\alpha^\ast dz$ where $z$ is a fixed coordinate on $\mathbb{C}$ and do the computations. This does not make sense a priori but may be defined after a map of Riemann surfaces is defined properly. 
