Let $\mathbb{C} \mathbb{P}^1 $ the complex projective space and let consider the tautological bundle over $\mathbb{C} \mathbb{P}^1 $:
In the excerpt below (whole document: here ) the map belonging to tautological bundle of $\mathbb{C} \mathbb{P}^1 $ is defined as the canonical map $s: \mathbb{C}^2 \backslash \{0\} \to \mathbb{C} \mathbb{P}^1 $ which assoziates every $(x_0:x_1)$ in canonical way the fiber $s^{-1}(x_0:x_1) = \mathbb{C}*(x_0,x_1)$, therefore the corresponding one dimensional vector space.
My question is how does this constuction yields $\mathcal{O}(-1)$ as line bundle, therefore a twisted sheaf with empty global sections (because of $-1$)?
Source:
First the map $s$ should from $\Bbb C^2 \backslash \{(0,0) \}$ to $\Bbb CP^1$. And this is not exactly this map which gives $\mathcal O(-1)$ as the fiber gives a punctured line for all $[x_0:x_1] \in \Bbb CP^1$.
Your reference didn't wrote the map precisely, so let's do it : the correct construction is to take the projection $X \to \Bbb CP^1$ where $X = \{(z,l) \in \Bbb C^2 \times \Bbb CP^1 : z \in l \}$. Here, the fiber is canonically a line as you said. This shows that $\mathcal O(-1)$ is a line bundle. For see that indeed it has no sections, the simplest thing is to write down the condition with the cocyles.
Remark : it is also interesting to see that the projection $X \to \Bbb C^2$ is the blow-up of $\Bbb C^2$ at the origin.