Find the quadratic variation process of $\int f(s) \, dB_s$

The assertion follows if we can show that

$$\lim_{\delta \to 0} \sup_{\|\Delta\| \leq \delta} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 = 0. \tag{1}$$

Recall the following result (see e.g. here or here)

Let $u \in L^1([a,b])$ be an integrable function. Then $u$ is uniformly integrable, i.e. for any $k \in \mathbb{N}$ there exists a constant $r>0$ such that $$\int_A |u(s)| \, ds \leq \frac{1}{k}$$ for all measurable sets $A \subseteq [a,b]$ with Lebesgue meausre $\leq r$.

Fix $k \in \mathbb{N}$. Since $u := f^2$ is integrable, we can choose $r>0$ such that $\int_A |f(s)|^2 \, ds \leq 1/k$ for any measurable set $A$ with Lebesgue measure $\leq r$. If $\Delta_n$ is a partition of $[a,t]$ with $\|\Delta_n\| \leq r$ we get

\begin{align*} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2&\leq \frac{1}{k} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right) \\ &= \frac{1}{k} \int_a^t f(s)^2 \, ds. \end{align*}

Hence,

$$\limsup_{\delta \to 0} \sup_{\|\Delta\| \leq \delta} \sum_{i=1}^n \left( \int_{t_{i-1}}^{t_i} f(s)^2 \, ds \right)^2 \leq \frac{1}{k},$$

and since $k \in \mathbb{N}$ is arbitrary this proves the assertion.

A final remark regarding your reasoning: To get the last equality in your computations I would rather use that $\int_u^v f(s) \, dB_s$ is Gaussian with mean zero and variance $\int_u^v f(s)^2 \, ds$ (.. note that this allows you to compute all moments of $\int_u^v f(s) \, dB_s$). There is no need to know the distribution of the squared integral.

You may be overthinking this. Because $f$ is square integrable, the function $g(u):=\int_a^u f(s)^2\,ds$ is continuous. Consequently, $$ \eqalign{ \sum_{i=1}^n\left(\int_{t_{i-1}}^{t_i} f(s)^2\,ds\right)^2 &\le\max_{1\le i\le n}[g(t_i)-g(t_{i-1}]\cdot \sum_{i=1}^n\int_{t_{i-1}}^{t_i} f(s)^2\,ds\cr &=\max_{1\le i\le n}[g(t_i)-g(t_{i-1}]\cdot \int_{a}^{t} f(s)^2\,ds\cr } $$ and the max above tends to $0$ as $n\to\infty$ because $g$ is uniformly continuous on $[a,t]$. This is all that's needed to finish your Chebyshev estimate argument.