Consider discrete time stochastic process $(X_0,\dots, X_i,\dots)$ adapted to filtration $\mathcal{F}_i$. Let $\tau$ be a stopping time. The following corollary is found in https://mathweb.ucsd.edu/~tkemp/280C/46.1.Strong-Markov-After.pdf on page 3.
Cor: $\mathcal{F}_\tau$ (Sigma algebra of stopping time) is independent of process $(X_{\tau+i})_{0\leq i}$ after conditioning upon $\tau$ finiteness and $X_\tau=x$.
Then on the last page of notes, it says $(X_0,\dots, X_\tau)$ is conditionally independent of $(X_\tau,X_{\tau+1},\dots)$ by conditioning upon $\tau$ finiteness and $X_\tau=x$.
$Q:$ Why is $(X_0,\dots, X_\tau)$ is even $\mathcal{F}_\tau$ measurable? $X_\tau$ is $\mathcal{F}_\tau$ measurable. Since $\sigma(X_0)\subset\mathcal{F}_0\subset\mathcal{F}_n$ for all $n$, it follows that $X_0\in\mathcal{F}_\tau$. Consider $B\in\sigma(X_1)\subset\mathcal{F}_1$. Suppose $B\not\in\mathcal{F}_0$. Then $(\tau=0)\cap B$ may not even inside $\mathcal{F}_0$ by definition of $\mathcal{F}_\tau$. In particular, $X_1$ might not even be $\mathcal{F}_\tau$ measurable.
$Q':$ Why it follows that $(X_0,\dots, X_\tau)$ is conditionally independent of $(X_\tau,X_{\tau+1},\dots)$ by conditioning upon $\tau$ finiteness and $X_\tau=x$?
Because $\tau$ is random, the r.v. $X_k$ in on the list $(X_0,\ldots,X_\tau)$ only if $\tau\ge k$.
One way to think about what the informal notation $(X_0,\ldots,X_\tau)$ means is to think in terms of the events determined by $X_0,\ldots,X_\tau$, restricted to $\{\tau<\infty\}$. These are the events of the form $$ \cup_{n=0}^\infty\{(X_0,\ldots,X_n)\in B_n, \tau=n\},\qquad(\dagger) $$ where $B_n$ is a measurable subset of $S^n$, for each $n$. As $\sigma(X_0,\ldots,X_n)\subset\mathcal F_n$, the event ($\dagger$) is an element of $\mathcal F_\tau$.