I'm trying to understand the computation of the canonical divisor of $\mathbb{P}^2$.
The usual explanation goes like this:
Let $(X:Y:Z)$ be the coordinates of $\mathbb{P}^2$. In the local chart $U_Z=\{Z\neq 0\}$, we define the coordinates $x:=\frac{X}{Z}$ and $y:=\frac{Y}{Z}$ and take the $2$-form $dx\wedge dy$, which has no zeros and no poles.
In $U_Y:=\{Y\neq 0\}$, we define $u:=\frac{X}{Y}$, $v:=\frac{Z}{Y}$, so that:
\begin{align*} dx\wedge dy&=d\left(\frac{u}{v}\right)\wedge d\left(\frac{1}{v}\right)\\ &=\left(\frac{du}{v}-\frac{u}{v^2}dv\right)\wedge\left(-\frac{1}{v^2}dv\right)\\ &=-\frac{1}{v^3}du\wedge dv \end{align*}
which has a pole of order $3$, therefore $K_{\mathbb{P}^2}=-3H$, where $H$ is a hyperplane (in this case, a line).
What I don't understand about this argumentation is 1) why don't we need to check the other chart $U_X:=\{X\neq 0\}$? and 2) what is this $H$ exactly?
1) $\mathbb{P}^2 \setminus (U_Z \cup U_Y)$ consists of the single point $(1:0:0)$. A divisor of the plane is a line, which can't be contained in this set, so we haven't missed anything.
2) $H$ is the hyperplane where there is a pole, namely $v=0$, i.e., $Z=0$. But all lines in $\mathbb{P}^2$ are linearly equivalent, so if you only care about the class in the Picard group, it doesn't matter which one you choose.