Suppose $F\in L_p(S_1\times S_2,\mu_1\otimes\mu_2)$. Then the mapping $f:S_1\rar:L_1(\mu_2)$, $f(x_1)\colon=F_{x_1}$ is measurable and if $p=1$ $f:S_1\rar L_1(S_2,\mu_2)$ is integrable and for $\mu_2$ almost all $x_2\in S_2$:
$$
\Big(\int_{S_1} f\,d\mu_1\Big)(x_2)
=\int_{S_1} F(x_1,x_2)\,\mu_1(dx_1)~.
$$
If $f:S_1\rar L_p(S_2,\mu_2)$ is integrable, then this relation also holds in $L_p(S_2,\mu_2)$.
As $(S_i,\F_i)$, $i=1,2$ are countably generated all spaces are separable. $f$ is measurable iff for all $g\in L_q(\mu_2)$ the mapping
$$
x_1\mapsto\int f(x_1)(x_2)g(x_2)\,\mu_2(dx_2)
=\int F_{x_1}g\,d\mu_2
$$
is measurable; but that's simply Fubini. For $p=1$ we have: $\int\norm f\,d\mu_1=\int|F|$ and thus $f$ is integrable; by Fubini
$$
f(x_2)\colon=\int F(x_1,x_2)\,\mu_1(dx_1)
$$
is in $L_1(\mu_2)$ and for all $g\in L_\infty(\mu_2)$ again by Fubini ($\la f,g\ra\colon=\int fg\,d\mu_2$):
\begin{eqnarray*}
\la\int f\,d\mu_1,g\ra
&=&\int\la f(x_1),g\ra\,\mu_1(dx_1)
=\int\int F(x_1,x_2)g(x_2)\,\mu_2(dx_2)\,\mu_1(dx_1)\\
&=&\int\Big(\int
F(x_1,x_2)\,\mu_1(dx_1)\Big)g(x_2)\,\mu_2(dx_2)
=\la f,g\ra~.
\end{eqnarray*}