The complement of $D$ in $\mathbb{P}^{1,\mathrm{an}}_K$ is a copy of the open unit disk $D^\circ$ with one additional rank $2$ point $x$ pointing 'towards the center' of $D^\circ$.
One can just do this 'by hand', but let me explain a more high-brow way of understanding this that more easily generalized to other situations.
Observe that $\mathbb{P}^{1,\mathrm{an}}_K$ has the natural formal model $\widehat{\mathbb{P}}^1_{\mathcal{O}_K}$. We then have the specialization map
$$\mathrm{sp}_{\mathbb{P}^{1}}\colon |\mathbb{P}^{1,\mathrm{an}}_K|\to |\widehat{\mathbb{P}}^1_{\mathcal{O}_K}|=|\mathbb{P}^1_k|,$$
where $k$ is the residue field of $K$. From this perspective (let me assume that $D_+$ (I'm using $+$ to indicate it is the north pole disk) is the closed unit disk centered at $0\in\mathbb{P}^{1,\mathrm{an}}_K(K)$) then $D_+=\mathrm{sp}_{\mathbb{P}^{1}}^{-1}(\mathrm{Spec}(k[t]))\subseteq \mathbb{P}^1_k$. So then, we see that $\mathbb{P}^{1,\mathrm{an}}_K-D_+=\mathrm{sp}_{\mathbb{P}^{1}}^{-1}(\infty)$.
We then see where the issue is coming from: $D_-^\circ$ is an open subset of $\mathbb{P}^{1,\mathrm{an}}_K$ but $\mathrm{sp}_{\mathbb{P}^1}^{-1}(\infty)$ is closed. In fact, $D_-^\circ =\mathrm{sp}^{-1}_{\mathbb{P}^1}(\infty)^\circ$ and is an example of a Berthelot tube: the inclusion $D_-^\circ\subseteq \mathbb{P}^{1,\mathrm{an}}_K$ is canonically identified with the generic fiber of the map $\mathrm{Spf}(\widehat{\mathcal{O}}_{\widehat{\mathbb{P}}^1_{\mathcal{O}_K},\infty})\to \widehat{\mathbb{P}}^1_{\mathcal{O}_K}$. In particular, we see that $\mathrm{sp}^{-1}_{\mathbb{P}^1}(\infty)-D_-^\circ=\partial_{\mathbb{P}^{1,\mathrm{an}}_K}D_+$.
So, we see that to understand the 'extra' points in $\mathbb{P}^{1,\mathrm{an}}_K-D_+$ come down to understanding the closure of $D_+$ in $\mathbb{P}^{1,\mathrm{an}}_K$. But, there is an easy way of doing this in general (at least topologically). For example, see Proposition 3.4.5 of my paper [ALY]. Namely, this shows that (topologically) $\overline{D}_+$ is the universal compactification of $D_+=\mathrm{Spa}(K\langle t\rangle)$ which (again by loc. cit.) is $\mathrm{Spa}(K\langle t\rangle, R)$ where $R$ is the integral closure of the $\mathcal{O}_K$-subalgebra of $\mathcal{O}_K\langle t\rangle$ generated by $K\langle t\rangle^{\circ\circ}=t\mathfrak{m}\langle t\rangle$, where $\mathfrak{m}\subseteq \mathcal{O}_K$ is the maximal ideal.
As it turns out, the complement of $D_+=\mathrm{Spa}(K\langle t\rangle)\subseteq \mathrm{Spa}(K\langle t\rangle,R)=\overline{D}_+$ is a single point (it is always finite in the curve case--see Corollary 3.4.6 of op. cit.): the valuation
$$v_{0,1^+}\colon K\langle t\rangle\to \mathbb{R}^{>0}\times(1^+)^{\mathbb{Z}},\qquad \sum_n a_n t^n\mapsto \sup_n(a_n \gamma^n),$$
where the ordering is such that $1<1^+<r$ where $r$ ranges over all $r\in\mathbb{R}$ with $r>1$. In other words, $1^+$ is a number 'infinitesimally larger than $1$'. Note that $v_{0,1^+}$ is not in $D_+=\mathrm{Spa}(K\langle t\rangle,\mathcal{O}_K\langle t\rangle$ itself as $|v_{0,1^+}(t)|>1$ so that $v_{0,1^+}(\mathcal{O}_K\langle t\rangle)$ is not bounded by $1$.All of the contents of this paragraph can be read off from the nice writeup in [Con]. Note that for all the Type 2 points $v_{x,r}$ of $D_+$ the closure is homeomorphic to $\mathbb{P}^1_k$ except at the Gauss point $v_{0,1}$ where the closure is homeomorphic to $\mathbb{A}^1_k$--from this perspective, the missing point $\infty\in\mathbb{P}^1_k-\mathbb{A}^1_k$ is this point $v_{0,1^+}$.
So, in summary, $\mathbb{P}^{1,\mathrm{an}}_K-D_+$ is $\mathrm{sp}_{\mathbb{P}^1}^{-1}(\infty)$ which differs from the Berthelot tube $\mathrm{sp}_{\mathbb{P}^1}^{-1}(\infty)^\circ=D_-^\circ$ of $\infty$ by a single point $v_{0,1^+}$ which is the missing 'outward facing' point of $D_+$, and points 'inward' towards $D_-^\circ$.
References"
[ALY] https://arxiv.org/pdf/2105.05184
[Con] https://math.stanford.edu/~conrad/Perfseminar/Notes/L11.pdf
