I understand the construction of $\mathcal{O}_{\Bbb{P}^1}(-1)$ as a sheaf on $\Bbb{P}_\Bbb{C}^1$, but I'm trying to understand how exactly does this define a line bundle and why people call this the "tautological line bundle".
Following the suggestion of "tautological", my first idea was to define the map: \begin{align*} \pi:\Bbb{A}^2&\to\Bbb{P}^1\\ (x_0,x_1)&\mapsto (x_0:x_1) \end{align*}
whose fibers are clearly lines.
Now, I can neither see how to define the trivialization maps nor how this relates to the sheaf $\mathcal{O}_{\Bbb{P}^1}(-1)$, so probably I'm on the wrong path.
I'm having a hard time trying to come up with different ideas, because I don't even know how to find a variety $X$ so that $\pi:X\to\Bbb{P}^1$ is the line bundle I'm looking for.
Maybe it helps to think about it if you don't choose a basis right away. Let $V$ be a two-dimensional vector space and $\mathbb{P}(V)$ the variety of lines through the origin in $V$. Then $\mathbb{P}(V)$ certainly has a trivial two-dimensional vector bundle $V \times \mathbb{P}(V)$. Inside this bundle is the more interesting bundle $$ \{ (v, \lambda) \in V \times \mathbb{P}(V): v \text{ is on the line } \lambda\}. $$ This is why it's called tautological: a point of $\mathbb{P}(V)$ is a line and this is the bundle that attaches that line to it.
Of course you can't compute like this; you need coordinates. In coordinates, what you want is $$ \{(x_0, y_0), [z_0: w_0] \in \mathbb{C}^2 \times \mathbb{P}^1 : \exists a \in \mathbb{C} \text{ such that } az_0 = x_0 \text{ and } aw_0 = y_0 \}. $$ together with its normal projection to $\mathbb{P}^1$.