Does ${\tau \wedge n}(ω)$ convergence uniformly to the random variable ${\tau}(ω)$ too?
My intuition is that, it does not, as a result of ${\tau \wedge n}(ω)$ converging to ${\tau}(ω)$ being dependent on $ω$, i.e the rate of convergence to the stopping time is not "uniform" for all $ω \in\ Ω $.
As a follow up question: Does the fact that ${\tau \wedge n}(ω)$ converges pointwise to the ${\tau}(ω)$, trivially implies that $X_{\tau \wedge n}\to X_{\tau}$ a.s, or are there subleties involved?
You are correct, the convergence is not uniform (unless $\tau$ is bounded, then $\tau \wedge n = \tau$ for all $n$ larger than the bound).
The main subtleties in $X_{\tau \wedge n} \rightarrow X_\tau$ are questions of whether $X_\tau$ and $X_{\tau \wedge n}$ are measurable and if $\tau$ can be infinite. If $X_t$ is progressively measurable then $X_\tau$ and $X_{\tau \wedge n}$ are measurable. If $\mathbb{P}(\tau = \infty) > 0$, then we need that there exists a random variable $X_\infty$ such that $\lim_{t \rightarrow \infty} X_t = X_\infty$ a.s. to have $X_{\tau \wedge n} \rightarrow X_\tau$.