Hint. Let $G$ be a graph which is the union of $k$ forests. (By the way, since a subgraph of a forest is a forest, each subgraph of $G$ is also the union of $k$ forests.)
(a) Find an upper bound (in terms of $k$ and $n$) for the size (number of edges) of $G$.
(b) Find an upper bound for the average degree of a vertex in $G$.
(c) Show that $\delta(G)\le2k-1$. ($\delta = $minimum degree.)
(d) Show that $\chi(G)\le2k.$ Use induction on the number of vertices, unless "degeneracy" has been covered. [P.S. Oops, I just noticed that you mentioned degeneracy in your question! I apologize for my careless reading!]
Spoilers:
An $n$-vertex forest has at most $n-1$ edges. Since $G$ has $n$ vertices and is the union of $k$ forests, $G$ has at most $k(n-1)$ edges. Hence the degree-sum of $G$ is at most $2k(n-1)$, and the average degree is at most $\dfrac{2k(n-1)}n=2k-\dfrac{2k}n\lt2k$. Since the average degree is less than $2k$, there must be a vertex of degree less than $2k$. Let $v$ be a vertex of degree less than $2k$. By the induction hypothesis, the graph $G-v$ is $2k$-colorable. Since $v$ is adjacent to at most $2k-1$ vertices in $G-v$, there is at least one color available for $v$.
This argument only applies to a finite graph $G$, but the theorem is also true for infinite graphs; the infinite case follows from the finite case by the De Bruijn–Erdős theorem.
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.)
Upper bound
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.)
Lower bound
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.
Absolute bounds
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
Your proof may be on the wrong track, since you can only utilise Euler's formula if the graph is connected, and I cannot think of how to make that work out. I provide a proof below which uses triangulations, which are connected and Euler's formula will apply.
This inequality is well known. For $n=1$ and $n=2$ it is obvious. First note that every planar graph with $v\ge 3$ vertices is a subgraph of a plane triangulation with the same vertex set (a planar graph where all faces have degree $3$). To see this, for every face with degree larger that $3$, we can choose one vertex $u$ on its boundary and add new edges from $u$ to every nonadjacent vertex on the boundary of the face; the resulting graph is a triangulation.
Let $e^\prime$ be the number of edges in the triangulation. Here, there are $3$ edges per face, so $3f=2e^\prime$ (each edge is adjacent to two faces). Then by Euler's formula, $$3e^\prime=3(v+f-2)=3v+2e^\prime-6.$$ Since your graph is a subgraph of the triangulation, $e\le e^\prime$; hence, after rearranging, $$e\le 3v-6.$$