There are many abuses of language in this subject and in order to help you I'll describe a few things completely rigorously.
Let $(U,\phi_\alpha)$ and $(V,\phi_\beta)$ be two charts (=local coordinates) of your Riemann surface $X$ at $P\in X$, that is $P\in U\cap V$.
The overlap is $\phi_\alpha(U\cap V)\subset \mathbb C$ and you have a holomorphic isomorphism $w: \phi_\alpha(U\cap V)\to \phi_\beta(U\cap V):z\mapsto w(z)$ which answers Question 1.
If you define $z_0=z(P)$, you can compute the derivative $w'(z_0)=\frac {dw}{dz}(z_0)\in \mathbb C$ and this answers Question 3.
Finally, since a holomorphic function $\mathbb \phi$ defined in a neighbourhood of $P$ has a differential which is a linear form $d\phi(P):T_P(X)\to \mathbb C$, you get two linear forms $d{\phi_\alpha} (P),d{\phi_\beta} (P):T_P(X)\to \mathbb C$.
Are they equal? Not at all! They are related by $$d{\phi_\beta} (P)=w'(z_0)\cdot d{\phi_\alpha}(P) \in T_P^*(X)$$
an equality in the fiber of the cotangent bundle at $P$ which answers (I hope!) your Question 2.
Over spheres, you actually have a nice way to distinguish bundles.
Every sphere $S^n$ can be written as a union of two contractible spaces, $U \cup V$, whose intersection is a small neighborhood of the equator, $\mathbb{R} \times S^{n-1}$. For instance if $x,y$ are the north and south poles, respectively, you can take $U = S^n - \{x\}, V = X^n - \{y\}$.
Over $U$ and $V$, any bundle is trivial, as you noted. The bundle on the sphere is glued together from these two trivial bundles.
If you're interested in a vector bundle, the transition function for the bundle is given by a map $U \cap V \to GL_m$, for a vector bundle of rank $m$. If this map is null-homotopic, meaning that it can be homotoped to the constant map, then your bundle is trivial, as any trivialization over $U$ can then be glued to a trivialization over $V$. So in the end you are reduced to the study of maps from $U \cap V \simeq S^{n-1}$ to your structure group -- i.e., $\pi_{n-1}GL_m$.
For $m=1$, this amounts to asking whether any map $S^{n-1} \to GL_1$ is null-homotopic. In your case for $S^3$, since $GL_1(\mathbb{R})$ is a disjoint union of two contractible spaces, and $S^2$ is connected, any map is null-homotopic.
In the fiber bundle case with fiber $S^1$ and base $S^3$, you are looking at maps $S^2 \to S^1$. But $\pi_2(S^1) = *$ so any circle bundle is contractible over $S^3$.
Note that both facts hold for higher spheres as well.
Best Answer
For simplicity let's take $P = [0: 1]$.
Cover $S$ with the two opens $U_0 = S \setminus [1: 0]$ and $U_\infty = S \setminus [0:1]$.
Now take the disjoint union of two copies of the trivial bundle: $M = (U_0 \times \mathbb{C}) \sqcup(U_\infty \times \mathbb{C})$
We impose an equivalence relation on $M$ by the rule $([a:b], z)_0 \sim ([a:b], bz/a)_\infty$ when neither $a$ nor $b$ are zero, and set $L = M/\sim$.
Since $\sim$ respects the projection to $S$, there is a natural map $L \to S$. The fiber over a point is $\mathbb{C}$, and local triviality is clear (every point is in $U_0$ or $U_\infty$).
To give a section of $L$ is to give a map $f_0: U_0 \to \mathbb{C}$ and a map $f_\infty: U_\infty \to \mathbb{C}$ such that $$ bf_0([a: b]) = af_\infty([a:b]) $$ whenever neither $a$ nor $b$ are zero.
There is such a section: $f_0([a: b]) = a/b$ and $f_\infty([a: b]) = 1$. This is locally holomorphic, so it's holomorphic, and it has a unique zero of order one at $P$ as desired.