If a map $f:Y\to X$ is an open surjection, then the inverse map $f^{-1}:X\to \mathcal K(Y)$ to the hyperspace $\mathcal K(Y)$ is continuous. The hyperpsace $\mathcal K(Y)$ is the space of nonpempty compact subsets endowed with the Vietoris topology.

So, your problem reduces to evaluating the Borel complexity of the set $\mathcal O:=\{K\in\mathcal K(Y): K\cap G$ is open in $K\}$ in $\mathcal K(X)$.

Fix a countable base $\mathcal B$ of the topology of $Y$, which is closed under finite unions. Observe that $\mathcal O=\{K\in\mathcal K(Y):\exists B\in\mathcal B\; B\cap K=G\cap K\}=\bigcup_{B\in\mathcal B}\mathcal O_B$ where $\mathcal O_B:=\{K\in\mathcal K(Y):B\cap K=G\cap K\}$.

Observe that $\mathcal O_B=\mathcal O_B'\cap\mathcal O_B''$ where $\mathcal O'_B=\{K\in\mathcal K(Y):B\cap K\subset G\}$ and $\mathcal O''_B=\{K\in\mathcal K(Y):G\cap K\subset B\}$.

If $K\notin\mathcal O'_B$, then $K$ intersects the open set $B\setminus G$ and $\{K'\in\mathcal K(Y):K'\cap(B\setminus G)\ne\emptyset\}$ is an open neighborhood of $K$ in $\mathcal K(Y)\setminus\mathcal O'_B$, which means that the set $\mathcal O'_B$ is closed in $\mathcal K(Y)$.

Observe that $K\in\mathcal O''_B$ is and only if $K\cap G\cap (X\setminus B)=\emptyset$, which means that the set $\mathcal O''_B$ is open in $\mathcal K(Y)$. So $\mathcal O_B$ is the intersection of an open and closed sets and $\mathcal O=\bigcup_{B\in\mathcal B}\mathcal O_B$ is an $F_\sigma$-set in $\mathcal K(Y)$ and so is its preimage $\{x\in X:f^{-1}(x)\cap G$ is open in $f^{-1}(x)\}$ in $X$.