Fields of meromorphic functions
Recall that the set of meromorphic functions $\mathcal{M}(X)$ over a Riemann surface $X$ is a field, under natural multiplication and addition of functions. Moreover, if $f:Y\rightarrow X$ is a holomorphic map, then $f$ induces a field homomorphism.
$\begin{aligned} f^*:\mathcal{M}(X)&\rightarrow \mathcal{M}(Y)\\ g&\mapsto g\circ f \end{aligned}$
In this post, we are going to present the relationship between fields of meromorphic functions, given such maps.
Lemma. Let $f:X\rightarrow Y$ be a nonconstant holomorphic map between Riemann surfaces. Then
$\begin{aligned} f^*:\mathcal{M}(Y)&\rightarrow \mathcal{M}(X)\\ \varphi&\mapsto\varphi\circ f \end{aligned}$
defines a field extension. In other words, $f^*\mathcal{M}(Y)\subset\mathcal{M}(X)$ is a subfield.
Proof. Consider $1_Y:Y\rightarrow\mathbb{C}$. Then $f^*(1_Y)=1_Y\circ\varphi=1_X$, which means that $1_X\in f^*\mathcal{M}(Y)$. Additionally, $f^*\mathcal{M}(Y)$ is closed under addition, multiplication and inverses. Thus $f^*\mathcal{M}(Y)$ is a subfield. $\square$
Field extension actually gives another possible kinds of degree. Now, we have the following notion of degrees
 degree of a covering map;
 degree of a polynomial (affine plane curve);
 degree of a field extension.
The applicable fact is that: these three kinds of curves coincide.
Here comes the main theorem of this post.
Theorem. Let $f:X\rightarrow Y$ be a nonconstant holomorphic branched covering of degree $d$.
$f^*:\mathcal{M}(Y)\hookrightarrow\mathcal{M}(X)$
Set $K=\mathcal{M}(Y)$. Then

$\forall g\in \mathcal{M}(X)$, the degree of field extension $[K(g):K]\leq d$;

If $[K(g):K]=d$, then $\mathcal{M}(X)=K(g)$;

If $g$ has $d$ distinct values on $f^{1}(b)$ for some $b\in Y$, then $[K(g):K]=d$.
Combining 2 and 3: if a meromorphic function $g\in\mathcal{M}(X)$ can be found, then the degree of field extension is $d$. In other words, $[\mathcal{M}(X):\mathcal{M}(Y)]=d$.
Proof. (1) Set $S=R(f)\cup\{p\in Xf(p)=\infty\}$ be the set of singular points in $X$, the subspace $Y'=Yf(S)$, and $X'=XS$. Then by the theorem in Covering, $f':X'\rightarrow Y'$ is unramified.
$\Rightarrow$ $\forall b\in Y'$, $\exists V\subset Y'$ an open neighborhood of $b$, s.t.
$f^{1}(V)=\bigsqcup_{i=1}^d U_i,$
where $\varphi_i:V\rightarrow U_i$ is a biholomorphic map. Let $g\in\mathcal{M}(X)$ and set
$g_i=g\circ\varphi_i:V\subset Y'\rightarrow\mathbb{C}$
Consider $\varphi:\mathcal{M}(Y)\rightarrow\mathbb{C}$, a polynomial with indeterminant $w\in\mathcal{M}(V)$, defined by
$\varphi(w)=\prod_{i=1}^d(wg_i)=:\sum_{i=0}^d b_iw^{d1}$
where $b_i\in\mathcal{O}(V)$, and $\varphi(f^*g)=0$. The function $b_i\in\mathcal{M}(V)$ can be extended to global meromorphic function $B_i\in\mathcal{M}(Y)$. Thus, $\varphi\in\mathcal{M}(Y)[w]$, and $\varphi(f^*g)=0$, which implies that $\varphi$ is a polynomial that vanishes $f^*g$. Since $\deg\varphi\leq d$, then
$[K(g):K]\leq d.$
This proves the first statement.
(2) Suppose that $[K(g):K]=d$. Let $h\in\mathcal{M}(X)$ be another element. Then there exists a primitive element $p\in\mathcal{M}(X)$, such that
$K\subset K(g)\subset K(g,h)=K(p)$
Then
$\begin{aligned} d&\geq [K(p):K]\\ &=[K(g,h):K(g)]\cdot[K(g):K]\\ &=[K(g,h):K(g)]\cdot d \end{aligned}$
which means that the equality must hold. In other words, $[K(g,h):K(g)]=1$. Therefore, $h\in K(g)$ and we are done for the second statement.
(3) Suppose that $g\in\mathcal{M}(X)$ and $b\in Y$ has distinct preimage under $f$. Let $g_i=g\circ\varphi_i$ as before. If $g_i(b)=g_j(b)$ for some $i\neq j$, then $g$ cannot be a solution to a polynomial equation of degree $d$. THERE IS STILL A GAP HERE!?!