differential of $X_t = g(t)\int_0^t f(s) dB_s$.
I haven't used ito Calculus over two year, I know what Ito formula is, and I remember we use it to solve SODE by applying it to some specific $f(t,X_t)$ so that in $df(t, X_t)$ there is no $dX_t$ term, only $dt$ and $dB_t$.
In this problem, what is the correct way of writing down $dX_t$? By product rule $$dX_t = g'(t)\left[ \int_0^t f(s) dBs\right] dt + g(t)\left[ \int_0^t f(s) dBs\right]' dt$$
I want to say the second term is $g(t) f(t)dB_t$, but I am not sure how to show this rigorously.
I assume $f(t)$ and $g(t)$ are deterministic functions and $B_t$ is a standard Brownian motion. The correct differential is \begin{align} d\left( g(t)\int_0^tf(s)dB_s\right)&=dg(t)\int_0^tf(s)dB_s+g(t)d\left(\int_0^tf(s)dB_s\right)\\ &=g'(t)dt\int_0^tf(s)dB_s+g(t)f(t)dB_t. \end{align} This is indeed an example of the product formula, which in its general form reads $d(X_tY_t)=dX_tY_t+X_tdY_t+dX_tdY_t$ with $X_t$ and $Y_t$ being generic Ito processes. In your example the third term is missing, as you probably know, because $g$ is deterministic and that term would be $\mathcal{O}(dt^{3/2})$ which is negligible compared to the other two terms as $dt\to0$.
Your example can be derived also from the Ito(-Doeblin) formula $$ df(t,X_t)=f_t(t,X_t)dt+f_x(t,X_t)dX_t+\frac12f_{xx}(t,X_t)(dX_t)^2, $$ where $X_t=\int_0^tf(s)dB_s$ and $f(t,x)=g(t)x$. So if you need a rigorous derivation you should just refer to the proof of the Ito formula and then just use it with this $f$ function.