graph theory
I have a lemma in my lecture notes as follow:
Lemma. If $G$ is a connected plane graph, then the dual of the dual
graph of $G$ is again $G$, or more precisely, $(G^*)^*=G$.
Why planarity of $G$ is a necessary condition? I mean, can we say "for a graph $G$ in a surface $S$ whose faces are simply connected, the dual of the dual of $G$ is isomorphic to $G$"?
Maybe this is something you want: Any graph $G$ can be embedded in a compact surface $S_g$ of genus $g$, for some $g$. The minimum of such $g$ is called the genus of $G$. Then faces of a non-planar graph may be defined as the regions of the embedding of $G$ in the $S_g$.
Let's look at bicubic planar graphs. As you say, there must be $2n$ vertices, $3n$ edges, and $n+2$ faces for some $n$. Suppose there are $F_j$ faces of length $j$; then we want lower and upper bounds on $F_4$ in terms of $n$.
Note that every 4-cycle in the graph must be a face in the planar embedding, because any chord in the 4-cycle would give you cycles of length 3, impossible in a bipartite graph. The graph you have at the end of the question is not a counterexample because it is not cubic (in a 3-regular graph, there would be two triangles on the bottom.)
The upper bound is $n+2$. This is obviously the maximum possible because it is the total number of faces, and it is achieved by the cube graph, where $n = 4$.
This is the only graph that achieves the bound, because it is the only cubic planar graph consisting only of squares. If all the faces are squares, then 4 times the number of faces must be 3 times the number of vertices, or $$ 4(n+2) = 6n $$ so $n = 4$.
If we exclude this single graph, then the upper bound is $n$, and graphs with $F_4 = n$ can be found for arbitrarily large $n$ (see the prism graph at the end of this answer.)
To prove this, we need to show that we cannot have $n + 1$ 4-sided faces and a single $j$-sided face, with $j > 4$. In this case $$ 6n = 4F_4 + jF_j = 4(n+1) + j $$ so $j = 2n - 4$. The same logic as this answer by Henning Makholm applies:
Now start by drawing the face with $2n - 4$ sides. That uses up $2n -4$ nodes. Since the graph is cubic, each of these nodes have an extra neighbor. But if those neighbors are all different, then there would be at least $4n - 8$ nodes in the graph which is too many.
Therefore, there must be a pair of corners of the $j$-gon that share a neighbor outside the $j$-gon. Since [all other faces are 4-cycles], this is only possible if these two nodes are separated by one corner in the $j$-gon. But then it's impossible for the extra leg of the corner between them to connect to anything, which is absurd. So the graph cannot exist.
(It's not that absurd to have an edge going to another graph component contained within a face, but such an edge would be a cut-edge, which cannot exist in a regular bipartite graph.)
The lower bound is an arbitrarily small multiple of $n$. We can see this by applying a bitruncation operation, which replaces every vertex by a hexagon and leaves a smaller version of each face (with the same number of sides). Thus, starting from the cube, we get a bitruncated cube (aka truncated octahedron) with 6 squares and 8 hexagons ($n = 12$).
Bitruncating that gives a polyhedron with 6 squares and 32 hexagons ($n = 36$).
Bitruncating that gives a polyhedron with 6 squares and 104 hexagons ($n = 108$).
Here are the corresponding planar graphs:
In general, bitruncating yields a graph with 3 times as many vertices and the same number of squares, so we can keep going and get $F_4$ to be less than $\epsilon n$ for any $\epsilon > 0$.
Another option would be a chamfering operation, which replaces every edge by a hexagon and every face by a smaller face of the same number of sides. Starting from the cube, this gives you 4 times as many vertices with each step, so $n$ grows even faster, while always keeping 6 squares.
As you mention, any 3-regular planar bipartite graph must have at least 6 squares (assuming you forbid digons, that is, two edges between the same vertices). One way to show this is that 3 times the number of vertices equals the sum of each face's length, so
$$\begin{align} 6n &= 4F_4 + 6F_6 + 8F_8 + \ldots \\ &\geq 4F_4 + 6(n+2-F_4) \\ &= 6n + 12 - 2F_4 \\ 2F_4 &\geq 12 \\ F_4 &\geq 6. \end{align}$$
(If you allow digons, you can get this guy with two squares: 
)
There is no absolute upper bound on $F_4$ (independent of $n$), since you can have prism graphs with arbitrarily many squares:
These all have $F_4 = n$ and are bipartite for any even $n$.
Best Answer
TravisJ is correct. The dual graph is defined for graphs in a surface, and depends on both the surface, and exactly how the graph is embedded into it.
For example, consider the "figure 8" graph consisting of a single vertex and two edges connecting it to itself. We can embed this graph in the torus in several ways:
You may be able to prove it more generally than for planar graphs, but the dual of a graph is not a property of the graph itself but rather of its embedding in a surface. So the theorem can only be stated in reference to surfaces.
Addendum I believe the modified question is true: if $G$ is a graph in some surface $S$ such that all the faces of $G$ are simply connected, then the dual of the dual of $G$ is isomorphic to $G$.
To see this, note that because the faces are simply-connected, they are homeomorphic to the unit disk. For each face, choose such a homeomorphism. Under this homeomorphism, the center of the disk is the dual vertex for the face, and the edges of the face are arc segments of the unit circle. The radii connecting the center to the midpoints of the various edges form curves breaking the face into smaller simply connected segments. Combining each of these curves with the corresponding curve in the face on the other side of the edge it leads to forms a dual edge. The dual edges and dual vertices form a representation of $G^*$. However for each vertex of $G$, it is surrounded by a finite number of those smaller simply connected segments, which combine together to form a dual face for the dual graph with the vertex inside it. Further, the portions of the original edges on this side of their midpoints connect the vertex to one point on each of the surrounding dual edges. Thus the original graph $G$ forms a representation of the dual of the graph $G^*$.
This argument (which I've only outlined) follows the same lines as the planar proof. Because of the assumption that the faces are simply connected, the topology of the surface does not get in the way of the proof.