I think the answer is no - take $X = Y = [0, 1]$, and $f(y, x) =y$. Pick some Borel set $E \subset [0, 1]$ such that $E$ and its complement have nonzero measure in every open interval, and let $\Sigma_0$ be the sigma algebra generated by the sets $[0, 1] \times E$ and $\mathcal B([0, 1]^2 \setminus [0, 1] \times E)$ where $\mathcal B$ denotes the restriction of the Borel sigma algebra to the given set.

Then $g(y, x) = \frac{1}{2}$ for $(y, x) \in [0, 1] \times E$ and $g(y, x) = y$ otherwise.

This has no modification that is continuous in $x$ for every $y$, since for every $y$ except for $y= \frac{1}{2},$ $g(y, \cdot)$ is essentially discontinuous everywhere.

Edit: The below answer is invalid, since $\Sigma_0$ is a sub sigma algebra of $Y$, not $X \times Y$.

I think the answer is no - take $X = Y = [0, 1]$, and $f(y, x) =y$. Pick some Borel set $E \subset [0, 1]$ such that $E$ and its complement have nonzero measure in every open interval, and let $\Sigma_0$ be the sigma algebra generated by the sets $[0, 1] \times E$ and $\mathcal B([0, 1]^2 \setminus [0, 1] \times E)$ where $\mathcal B$ denotes the restriction of the Borel sigma algebra to the given set.

Then $g(y, x) = \frac{1}{2}$ for $(y, x) \in [0, 1] \times E$ and $g(y, x) = y$ otherwise.

This has no modification that is continuous in $x$ for every $y$, since for every $y$ except for $y= \frac{1}{2},$ $g(y, \cdot)$ is essentially discontinuous everywhere.

I think the answer is no - take $X = Y = [0, 1]$, and $f(y, x) =y$. Pick some Borel set $E \subset [0, 1]$ such that $E$ and its complement have nonzero measure in every open interval, and let $\Sigma_0$ be the sigma algebra generated by the sets $[0, 1] \times E$ and $\mathcal B([0, 1]^2 \setminus [0, 1] \times E)$ where $\mathcal B$ denotes the restriction of the Borel sigma algebra to the given set.

Then $g(y, x) = \frac{1}{2}$ for $(y, x) \in [0, 1] \times E$ and $g(y, x) = y$ otherwise.

This has no modification that is continuous in $x$ for every $y$, since for every $y$, $g(y, \cdot)$ is essentially discontinuous everywhere except at $x = \frac{1}{2}$.

I think the answer is no - take $X = Y = [0, 1]$, and $f(y, x) =y$. Pick some Borel set $E \subset [0, 1]$ such that $E$ and its complement have nonzero measure in every open interval, and let $\Sigma_0$ be the sigma algebra generated by the sets $[0, 1] \times E$ and $\mathcal B([0, 1]^2 \setminus [0, 1] \times E)$ where $\mathcal B$ denotes the restriction of the Borel sigma algebra to the given set.

Then $g(y, x) = \frac{1}{2}$ for $(y, x) \in [0, 1] \times E$ and $g(y, x) = y$ otherwise.

This has no modification that is continuous in $x$ for every $y$, since for every $y$ except for $y= \frac{1}{2},$ $g(y, \cdot)$ is essentially discontinuous everywhere.