Let $X=\mathbb{R}/\mathbb{Z}$. If $f\in L^2(X)$, we have the Fourier series expansion $f(x)=\sum_{k=-\infty}^\infty c_ke^{2i\pi k(x)}$ which converges in $L^2$.
Now, suppose $Y$ is an arbitrary probability space and let $F\in L^2(X\times Y)$. We have $F(\cdot,y)\in L^2(X)$ for a.e. $y$, and so we can write $F(x,y)=\sum_{k=-\infty}^\infty c_k(y)e^{2i\pi k(x)}$.
Is it generally the case that $c_k\in L^2(Y)$?
Yes, since the Fourier coefficients are given by inner products with the pure exponentials, which I will denote by $e_k$, so $e_k(x) = e^{2\pi i k x}$, and hence by Cauchy Schwarz $$ \int |c_k(y)|^2 dy = \int |\langle F(\cdot,y), e_k\rangle|^2 dy \leq \int \|F(\cdot,y)\|_{L^2}^2 \|e_k\|_{L^2}^2 d y = \|F\|_{L^2}^2 \cdot \|e_k\|_{L^2}^2. $$ Here, Tonelli's theorem was used in the last step.
Note: with a modification of this argument (using the Parseval identity) you can even show that $\int \sum_k |c_k(y)|^2 dy <\infty$.