No, this is not always possible. Indeed, the one-point compactification of a Riemann surface may not even be a manifold. For instance, the one-point compactification of $\mathbb{C}\setminus\{0\}$ is a sphere with two points identified, which is not locally homeomorphic to $\mathbb{R}^2$ at that point.
If $S$ and $T$ are Riemann surfaces which happen to have one-point compactifications $S^*$ and $T^*$ which are Riemann surfaces, then any proper holomorphic map $f:S\to T$ does extend holomorphically to a map $S^*\to T^*$. By properness, we can extend $f$ continuously to a map $S^*\to T^*$, and this map is holomorphic at $\infty$ since the singularity must be removable (in coordinate charts at $\infty$, $f$ is bounded near the singularity).
Yes. This is known as the divisor - line bundle correspondence, see section 2.3 of Huybrechts' Complex Geometry: An Introduction for example.
A divisor in a complex manifold $Y$ is a locally finite, formal, integral linear combination of irreducible hypersurfaces of $Y$. That is, a divisor takes the form $\sum_Z \eta_ZZ$ where $\eta_Z \in \mathbb{Z}$, the sum is taken over all irreducible hypersurfaces $Z \subset Y$, and the collection $\{Z \mid \eta_Z \neq 0\}$ is locally finite. We say a divisor is effective if $\eta_Z \geq 0$ for all $Z$. An analytic subvariety of codimension one is therefore an example of an effective divisor.
On a complex manifold, one can construct a holomorphic line bundle $\mathcal{O}(D)$ associated to a divisor $D$, which admits a meromorphic section $\sigma$ whose associated divisor is $D$. If the divisor was effective, then the section $\sigma$ is actually holomorphic. In particular, the line bundle associated to an analytic subvariety $Z$ of codimension one admits a holomorphic section $\sigma$ such that $\sigma^{-1}(0) = Z$.
Beyond complex manifolds, you must be careful. There are two notions of divisors in algebraic geometry, namely Weil and Cartier; they do not coincide in general. The definition of divisor I gave above is actually the definition of a Weil divisor, while the notion of Cartier divisors is the one which gives rise to holomorphic line bundles with sections.
Best Answer
You have to know some (basic) facts about complex/holomorphic line bundles over complex manifolds. I'll try to be much clear as possible.
(1) The first Chern class is a topological invariant for complex line bundles in the sense that it gives a group homomorphism $c_1:H^1(X,\mathcal{O}^*_X)\to H^2(X,\mathbb{Z})$. So we can think of $H^2(X,\mathbb{Z})$ as the group parametrizing complex line bundles.
(2) Holomorphic line bundles are parametrized by $H^1(X,\mathcal{O}^*_X)$
(3) There exists a long exact sequence of cohomology group, which the relevant part (for our purpose) is the following \begin{equation*}\cdots\to H^1(X,\mathcal{O}_X)\to H^1(X,\mathcal{O}_X^*)\overset{c_1}{\to}H^2(X,\mathbb{Z})\to H^2(X,\mathcal{O}_X)\to\cdots \end{equation*}
But $H^2(X,\mathcal{O}_X)\cong H^{0,2}_{\bar{\partial}}(X)$ and since we are on a Riemann surface there no exist $(0,2)$-forms and therefore $H^{0,2}_{\bar{\partial}}(X)\cong 0$. In particular from this follows that $c_1$ is surjective and by (1),(2) above that any complex line bundle over a Riemann surface admits a holomorphic structure.
For your second question just note that $H^2(X,\mathbb{Z})\cong\mathbb{Z}$, so you can take your favorite integer $n$ and find a complex line bundle $L$ such that $\int_Xc_1(L)=n$. For example if $n>0$ take $\tilde{L}$ to be a degree $1$ line bundle and define $L:=\tilde{L}\underbrace{\otimes\dots\otimes}_{n \ \text{times}}\tilde{L}$ (note that $\mathbb{Z}$ is generated by $1$ and $-1$ as additive group)