We look at the integral at time $t+dt$:
\begin{equation}
\begin{split}
\int_0^{t+dt}f(s,t+dt)dW_s &= \int_0^{t+dt}\left[f(s,t)+\partial_t f(s,t)dt \right]dW_s \\
&= \int_0^{t}\left[f(s,t)+\partial_t f(s,t)dt \right]dW_s+\left[f(t,t)+\partial_t f(t,t)dt \right]dW_t
\end{split}
\end{equation}
In the first line we expanded the integrand around $t$ in the second argument of $f$. In the second line we approximated the integral from $t$ to $t+dt$ by its integrand evaluated at $t$. Now we may use the rules of stochastic calculus to drop the term $dt\, dW_t$ and write
\begin{equation}
\begin{split}
\int_0^{t+dt}f(s,t+dt)dW_s &= \int_0^{t}f(s,t)dW_s + dt\cdot \int_0^{t}\partial_tf(s,t)dW_s + f(t,t)dW_t.
\end{split}
\end{equation}
Thus, if we define
\begin{equation}
\begin{split}
X^{(0)}_t=\int_0^{t}f(s,t)dW_s, \quad X^{(n)}_t=\int_0^{t}f^{(0,n)}dW_s,
\end{split}
\end{equation}
where $f^{(0,n)}$ is the $n$'th derivative of $f$ in the second argument, $X^{(0)}_t$ satisfies the Ito SDE
\begin{equation}
\begin{split}
dX^{(0)}_t=X^{(1)}_t dt+f(t,t)dW_t.
\end{split}
\end{equation}
One remark: In practice this SDE is not very helpful on its own, since $X^{(1)}$ is of the same form as $X^{(0)}$ and its SDE depends on $X^{(2)}$, and so on. This generates a closure problem, requiring us to always know one of the integrals beforehand. However, if $f^{(0,n)}\equiv 0$ for some positive integer $n$, we get a closed system of SDEs
\begin{align}
dX^{(0)}_t&=X^{(1)}_t dt+f(t,t)dW_t,\\
dX^{(1)}_t&=X^{(2)}_t dt+f^{(0,1)}(t,t)dW_t,\\
\vdots\quad & \qquad\qquad\vdots\\
dX^{(n-2)}_t&=X^{(n-1)}_t dt+f^{(0,n-2)}(t,t)dW_t,\\
dX^{(n-1)}_t&=f^{(0,n-1)}(t,t)dW_t.
\end{align}
Based on saz' comments:
'if we integrate over the whole space, then we don't have to change the bounds of integration'
$$\int_{\mathbb R} \int_{\mathbb R} f(x,y) dx dy = \int_{\mathbb R} \int_{\mathbb R} f(x,y) dy dx$$
for applicable f. Also,
$$\int_{\Omega} \int_{\mathbb R} f(t,\omega) dt d\mathbb P = \int_{\mathbb R} \int_{\Omega} f(\omega,t) d\mathbb P dt$$
for applicable f. Now apply: $$f(t, \omega) := X_t^2(\omega)1_{[0,T]}(t)$$
A notable difference b/w the 2 Fubini's Thms is that in basic calculus, f is required to be continuous. In stochastic calculus, $f$ is not required to be continuous, but $f$ is required to be measurable.
Best Answer
I will mention a Fubini-type theorem for double integrals involving a stochastic integral and a Lebesgue integral. But keep in mind that there are results for double stochastic integrals as well.
Let $g: X\times[0,\infty)\times\Omega \mapsto \mathbb{R}$ be a parametrized stochastic process with the parameter taking values in $X$ and $\mu$ be a $\sigma$-finite measure on a $\sigma$-algebra defined on this space. $g$ satisfies certain measurability properties on the triple product $\sigma$-algebra, which I won't get into as this is a very technical area.
The example you gave does not fit into this framework as the outer integral on the left hand side is on $\Omega$, not a separate parameter space. You cannot treat it as a parameter space as it is coupled to the inner integral. I don't see how you can exchange two integrals that take place on the same domain in different ways. Well, technically the stochastic integral is on $[0,t]$ but it is defined using convergence in $\Omega$.