Here is an example from the book of differential analysis on compact manifold. But I feel confused with using the transition function $g_{0,1}^k$ to define the line bundle $E^k\to P_1(\mathbb{C})$. Can anyone explain about this?
Here $U_{r,n}$ is the disjoint union of the $r$-planes ($r$-dimensional $\mathbb{C}$-linear subspaces) in $\mathbb{C}^n$.
If vector bundles $E$ and $F$ have transition functions $g_{\alpha\beta}$ and $h_{\alpha\beta}$, then $E\otimes F$ has transition functions $g_{\alpha\beta}h_{\alpha\beta}$. Therefore the bundle $E^k$ has transition functions $g^k_{\alpha\beta}.$
And to see why the transition functions for the tautological line bundle $E$ are what they are, let $(\ell,v)$ be a point in $E$. So $v\in\ell.$ In homogeneous coordinates, $\ell=[z_0,z_1]$ is a line in $\mathbb{C}^2$ spanned by the vector $(z_0,z_1).$ So $v=k(z_0,z_1)$ for some $k\neq 0.$
So in the affine patch $U_1 = \{[z_0,z_1]|z_0\neq0\},$ with coordinate $z=z_1/z_0$ we have a nowhere vanishing section $e_1\colon\ell\mapsto (1,z)$, so $v=kz_0e_1$ in the induced trivialization. And in affine patch $U_1 = \{[z_0,z_1]|z_1\neq0\},$ with coordinate $w=z_0/z_1$, our non-vanishing section is $e_2\colon\ell\mapsto (w,1).$ Under this trivialization, we have $v=kz_1e_2.$ Thus the transition from the $U_0$ to the $U_1$, the transition function is $(z_0/z_1)$.
So $E^k$ has transition functions $(z_0/z_1)^k.$