Finite morphism induced by section of a line bundle Assume $X$ is smooth, irreducible curve (curve = $1$-dimensional proper scheme over a fixed base field $k$)  and $x \in X$ a closed point. Since $\{x\}$ is a divisor of $X$, we obtain a line bundle $\mathcal{L}:= O_X(\{x\})$ a.k.a. invertible sheaf on $X$. We assume that  $\dim_k H^0(\mathcal{L}, X) \ge 2$. Therefore $H^0(\mathcal{L}, X)$ contains a non constant section $f \in H^0(\mathcal{L}, X)$. $f$ we can also see as an element of fraction field $K(X)$ of $X$ and therefore we can talk about $(f),(1/f)$ as $O_X$-modules.
Let $U_f:=X \backslash \text{Supp}(f)$ and $U_{1/f}:=X \backslash \text{Supp}(1/f)$. We obtain a cover $X=U_f \cup U_{1/f}$ of $X$ and can define a well defined map $f: X \to \mathbb{P}^1 _k= \text{Proj}(k[T_0,T_1])$ as follows on affine pieces:
$U_f \to D_+(T_0)$ corresponds to $T_1/T_0 \mapsto f$ and $U_{1/f} \to D_+(T_1)$ corresponds to $T_0/T_1 \mapsto 1/f$. It's easy exercise to verify that these maps glue to a map $f: X \to \mathbb{P}^1 _k$. By construction $f$ is affine and $f^{-1}(\infty)=x$, because $f$ is non constant and $\mathcal{L}= O_X(\{x\})$ (recall,  $\infty =(0:1) \in \mathbb{P}^1$).
My question is why is $f$ a finite morphism? Since it's a local property and $f$ is defined on affine pieces symmetrically, that suffice to understand why is $k[T_1/T_0] \to O_X(U_f),  T_1/T_0 \mapsto f$ is a finite map or equivalently why is  $O_X(U_f)$ a finite $k[T_1/T_0]$-module?
 A: This was intended as a comment, but it is too long to fit in.
I think in this case $\mathcal L$ is an ample line bundle of $X$, and the morphism $f$ in question is the canonical morphism (stacks project Lemma 01PZ).
Since the canonical morphism is an open immersion (stacks project Lemma 01Q1), the morphism $k[T_1/T_0]\rightarrow\mathcal O_X(U_f)$ should be a localization. Since being finite is a local property (stacks project Lemma 01WI), we may assume it is the localization at one element.
Since $X$ is a curve over $k$, that element should be finite over $k$ (as it is proper over $k$), i.e. satisfy a polynomial equation with coefficients in $k$. This shows that the localization in question is finite. 

Perhaps there is a simpler way to show this, but I hope this could still be of some use.

Please point it out if some errors occur or if there are any inappropriate points, thanks.
EDIT:
$(1):$
Consider a ring extension $A\subseteq B$ and $x\in B$ which is finite over $A$. Then $A_x$ is finite over $A$. This is because $x$ being finite over $A$ implies $x$ is integral over $A$, and hence $x$ satisfies a monic equation $x^n+\sum_{i=0}^{n-1}a_ix^i=0$. Dividing by $x^n$, we have
$$1+\sum_{i=0}^{n-1}a_ix^{i-n}=0.$$
As a consequence $A_x/A$ is generated by $\{1/x^k\mid k = 0, \cdots, n-1\}$ as a module.
As to why $x$ is finite, see this MO question, or this blog post.
$(2):$
Not any localization is the localization at one element: For example, consider the localization $\mathbb Z_{(2)}$ of $\mathbb Z$ at the prime ideal $(2)$: it consists of elements of the form $\frac ab$, where $a,b\in\mathbb Z$ and $2\not\mid b$. There is no element $x$ such that $\mathbb Z_{(2)}=\mathbb Z_x$ (otherwise $x$ has infinitely many prime divisors).
On the other hand, I think my original argument involving PID is wrong, so I edited the above argument as well. Thanks for pointing out the flaw.
