I know that: $$ \mathbb{E} \left[ \int_0^af(t)dB(t)\int_a^bf(t)dB(t) \right] = \mathbb{E} \left[ \int_0^af(t)dB(t) \right] \mathbb{E} \left[ \int_a^bf(t)dB(t) \right] = 0,$$ where $0<a<b$, because Brownian motions don't 'overlap' and are independent.
However, I can not find no credible source where this fact is stated or proven. Can you recommend where could I find this fact or theorems which lead to this fact, something like: $$\mathbb{E}\int_0^af(t)dB(t)\int_0^bf(t)dB(t)=\mathbb{E}\left(\int_0^{\min(a,b)}f(t)dB(t)\right)^2=\int_0^{\min(a,b)}f^2(t)dt.$$
Let's consider a more general fact:
Proposition Let $X_t$ and $Y_t$ be two adapted processes such that for a deterministic time $T > 0$ we have $\int_0^T E[X^2(t)]dt < \infty$ and $\int_0^T E[Y^2(t)]dt < \infty$. Then $$E \left[ \int_0^T X_t dB_t \int_0^T Y_t dB_t \right] = \int_0^T E[X_t Y_t] dt$$
Using this proposition, and assuming $f \in L^2$ and $a <b$, we have: $$\begin{align*} E \left[ \int_0^a f(t) dB_t \int_a^b f(t) dB_t \right] &= E \left[ \int_0^b f(t) 1_{\{ t \le a \}}dB_t \int_0^b f(t) 1_{ \{a \le t \le b\} } dB_t \right] \\ &= \int_0^b E[f^2(t) 1_{\{ t \le a \}}1_{ \{a \le t \le b\} }] dt = 0 \end{align*}$$