Understand better stochastic integral through a.s. convergence

The convergence of the subsequence holds almost surely, i.e. there is an exceptional null set where convergence fails to hold. This null set depends on the partitioning sequence $(t^{(n)})_{n \geq 1}$. Since there are uncountably many sequences, this is pretty bad - the union of the null exceptional sets is going to be quite huge, in general, and therefore we cannot expect to have a "uniform" exceptional null set for all sequences $(t^{(n)})_{n \in \mathbb{N}}$ with mesh size converging to zero. In fact, that's exactly the reason why the Itô integral is defined as an $L^2$-limit of the Riemann sums and not as a pointwise limit.

The phenomena which you are observing is a very general one: If a sequence of random variables $Y_n$ converges to a random variable $Y$ in probability (or in $L^2$), then we can choose a subsquence which converges almost surely to $Y$. Nevertheless, convergence in probability is a much weaker notion of convergence than pointwise convergence; this means, in particular, that the pointwise convergence of a subsequence of Riemann sums is far from giving a notion for a pointwise integration: If we want to get a pointwise notion for a stochastic integral, then we would like to fix $\omega \in \Omega$ and then define the stochastic integral, say, as a pointwise limit of Riemann sums along a suitable partition. That's, however, not what happens if we use the subsequence procedure from your question. If we take a sequence $(t^{(n)})_n$ then we get pointwise convergence with probability $1$ but we have no control about the null set. In particular, we don't have a clue how to choose a sequence $(t^{(n)})_n$ such that the Riemann sums convergence for our fixed $\omega$.

Let me give one further remark. In order to get the convergence

$$\int_0^T f(s) \, dB_s = \lim_{n \to \infty} \sum_{i=1}^n f(t_i^{(n)}) (B_{t_{i+1}^{(n)}}-B_{t_i^{(n)}}) \quad \text{in $L^2$}$$

(and hence the pointwise convergence of the subsequence of Riemann sums) you will typically need some continuity assumptions on $f$. If $f$ is a general progressively measurable function with $\mathbb{E}\int_0^t f(s)^2 \, ds < \infty$ for all $T>0$, then there exists some sequence of approximating simple functions $(f_n)_{n \in \mathbb{N}}$, i.e. a sequence of functions such that$$\mathbb{E}\int_0^t |f_n(s)-f(s)|^2 \ ds \to 0 \quad \text{and} \quad \int_0^t f(s) \, dB_s = L^2-\lim_{n \to \infty} \int_0^t f_n(s) \, dB_s;$$

however, the approximating functions $f_n$ will be, in general, not of the form

$$f_n(s) := \sum_{i=1}^n f(t_i^{(n)}) 1_{[t_i^{(n)},t_{i+1}^{(n)})}(s)$$

(which would give rise to the Riemann sums you are stating at the very beginning of the question); see e.g. Proposition 15.16 and Lemma 15.19/Theorem 15.20 in the book by Schilling & Partzsch (2nd edition) for more information.