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$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar.

This is definitely not a complete characterization: It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$. For instance, if $\mathcal{G}_A=K_4$, then $\mathcal{G}_B$ is a cube, and is planar, but the embedding derived from the planar embedding of $\mathcal{G}_A$ is a toroidal embedding of the cube with four hexagonal faces (the four equators of a standard cube) instead of the usual planar embedding with six square faces.

Some planar graphs $\mathcal{G}_A$ will definitely give nonplanar graphs $\mathcal{G}_B$. For instance, if $\mathcal{G}_A$ is the graph of the regular dodecahedron, its bipartite double cover is the cubic symmetric graph on 40 vertices, which is nonplanar. In the other direction, it is also possible for $\mathcal{G}_B$ to be planar even when $\mathcal{G}_A$ is nonplanar; for instance, if $\mathcal{G}_A$ is the planar graph formed from $K_{3,3}$ by subdividing the edges of a 6-cycle, then $\mathcal{G}_B$ is the planar graph of the hexagonal prism (with both of its 6-cycles subdivided in the same way).

$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar.

This is not a complete characterization: It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$. For instance, if $\mathcal{G}_A=K_4$, then $\mathcal{G}_B$ is a cube, and is planar, but the embedding derived from the planar embedding of $\mathcal{G}_A$ is a toroidal embedding of the cube with four hexagonal faces (the four equators of a standard cube) instead of the usual planar embedding with six square faces.

Some planar graphs $\mathcal{G}_A$ will definitely give nonplanar graphs $\mathcal{G}_B$. For instance, if $\mathcal{G}_A$ is the graph of the regular dodecahedron, its bipartite double cover is the cubic symmetric graph on 40 vertices, which is nonplanar. In the other direction, it is also possible for $\mathcal{G}_B$ to be planar even when $\mathcal{G}_A$ is nonplanar; for instance, if $\mathcal{G}_A$ is the nonplanar graph formed from $K_{3,3}$ by subdividing the edges of a 6-cycle, then $\mathcal{G}_B$ is the planar graph of the hexagonal prism (with both of its 6-cycles subdivided in the same way).

$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar.

This is definitely not a complete characterization: It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$. For instance, if $\mathcal{G}_A=K_4$, then $\mathcal{G}_B$ is a cube, and is planar, but the embedding derived from the planar embedding of $\mathcal{G}_A$ is a toroidal embedding of the cube with four hexagonal faces (the four equators of a standard cube) instead of the usual planar embedding with six square faces.

$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar.

This is definitely not a complete characterization: It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$. For instance, if $\mathcal{G}_A=K_4$, then $\mathcal{G}_B$ is a cube, and is planar, but the embedding derived from the planar embedding of $\mathcal{G}_A$ is a toroidal embedding of the cube with four hexagonal faces (the four equators of a standard cube) instead of the usual planar embedding with six square faces.

Some planar graphs $\mathcal{G}_A$ will definitely give nonplanar graphs $\mathcal{G}_B$. For instance, if $\mathcal{G}_A$ is the graph of the regular dodecahedron, its bipartite double cover is the cubic symmetric graph on 40 vertices, which is nonplanar. In the other direction, it is also possible for $\mathcal{G}_B$ to be planar even when $\mathcal{G}_A$ is nonplanar; for instance, if $\mathcal{G}_A$ is the planar graph formed from $K_{3,3}$ by subdividing the edges of a 6-cycle, then $\mathcal{G}_B$ is the planar graph of the hexagonal prism (with both of its 6-cycles subdivided in the same way).

$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar. (It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$, I'm not sure.)

$\mathcal{G}_B$ is the bipartite double cover of $\mathcal{G}_A$. So it has twice as many vertices and edges, and if $\mathcal{G}_A$ has an embedding with $x$ even faces and $y$ odd faces then $\mathcal{G}_B$ has an embedding (possibly on a nonplanar surface) with $2x+y$ faces, formed by making two copies of each even face and replacing each odd face by its double. Therefore, by Euler's formula, when the embedding of $\mathcal{G}_A$ is planar, the genus of $\mathcal{G}_B$ is at most $(y-2)/2$. In particular, if $\mathcal{G}_A$ is already bipartite, $\mathcal{G}_B$ consists of two disjoint copies of $\mathcal{G}_A$, and if $\mathcal{G}_A$ has only two odd faces then $\mathcal{G}_B$ will still be planar.

This is definitely not a complete characterization: It might also be possible for $\mathcal{G}_B$ to be planar in some other cases, by using a different embedding from the one derived from $\mathcal{G}_A$. For instance, if $\mathcal{G}_A=K_4$, then $\mathcal{G}_B$ is a cube, and is planar, but the embedding derived from the planar embedding of $\mathcal{G}_A$ is a toroidal embedding of the cube with four hexagonal faces (the four equators of a standard cube) instead of the usual planar embedding with six square faces.