Let $K$ be a non-Archimedian local field. We can define $\mathbb{P}^1_K$ in the standard way and the blow up $$\text{Bl}_0 K^2=\left\{ ((x_1,x_2),[y_1,y_2])\in K^2\times \mathbb{P}^1_K \lvert x_1y_2=x_2y_1\right\}$$ also in the standard way.
What is not standard is that $\mathbb{P}^1_K$ can be covered by disjoint compact-opens $U=\left\{[u:v]\lvert |u/v|\leq 1\right\}$ and $V=\left\{[u:v]\lvert |v/u|<1\right \}$. Hence we can write the blow up as the disjoint union $$\text{Bl}_0K^2 = (\text{Bl}_0K^2\cap (K^2\times U))\sqcup (\text{Bl}_0K^2\cap (K^2\times V))$$ Call these sets $\tilde{U}$ and $\tilde{V}$.
I have read in two different sources that $\tilde{U}$ and $\tilde{V}$ are both bi-analytic to $K^2$ and in these coordinates the projection maps to $K^2$ become $(s,t)\mapsto (s,st)$ and $(a,b) \mapsto (ab,a)$. The only way to make sense of this is that the bi-analytic map $\phi:K^2\to \tilde{U}$ sends $(s,t)$ to $(s,st)$ in the first two coordinates. Looking at the equations defining the blowup, we see what $(s,t)$ gets sent to in projective coordinates, namely $[1:t]$. That is, I must infer that this map $\phi$ sends $(s,t)$ to $((s,st),[1:t])$
However, I think this is incorrect. The projective coordinates $[u:v]$ in $\tilde{U}$ must satisfy $|u/v|\leq1$. For anything in the image of $\phi$, the projective coordinates are of the form $[1:t]$ and it is not true that $|1/t|\leq1$ for arbitrary $t$. I have come up with the following maps instead:
$$\phi: K\times \mathcal{ O}_K \to \tilde{U} \text{ given by } (a,b)\mapsto ((ab,a),[b:1])$$ $$\psi: K\times \pi\mathcal{ O}_K \to \tilde{V} \text{ given by } (s,t)\mapsto ((s,st),[1:t]) $$
$\mathcal{O_K}$ is the ring of integers of $K$ and $\pi$ is a uniformizer (generator of the unique nonzero prime ideal).
I haven't written out the details completely but I am confident these are bi-analytic maps (certainly they are analytic and their inverses are essentially just projection).
So...what gives? Did I make a mistake somewhere? Or did the authors of the notes I am reading make a mistake when they said these opens are bi-analytic to the full $K^2$?