Lattice Walk on Diagonally Overlapping Square Lattices Having worked on this very interesting question here, I wondered if there might be a general form for the number of paths for a lattice walk on diagonally overlapping square lattices from one end of the main diagonal to the other.
We define the following variables. 
Each square lattice has dimensions $n\times n$ single squares. 
There are $m$ square lattices, which overlap along a common diagonal. The overlapping area is a square lattice of $q\times q$ single squares, where $0\leq q<n$. We refer to this as an overlap of degree $q$. As an illustration, the diagram below shows the configuration for $n=5, m=4, q=2$. 
$\hspace{4cm}$
The preliminary conclusion is that, for the following cases, the solutions appear to have fairly neat closed forms. 
(i) Overlap of $\bf0$ ($q=0$)
This is a trival case where the square lattices are joined at the tip with no overlapping squares. Number of paths is 
$$\binom {2n}n^m$$
(ii) Overlap of degree $\bf1$ ($q=1$)
This is the case described in the question  referred to in the first paragraph. As shown in my solution, the general form for the number of paths is 
$$2\binom {2n-1}{n-1}^m=2\binom{2n-1}n ^m$$
(iii) Overlap of degree $\bf {(n-1)}$ ($q=n-1$)
This is a one-square-apart zig-zag formation. The problem may be simplified by considering it as a truncated ${N}\times {N}$ lattice (where $N=n+m-1$) with truncation point at $\pm n$ from the centre of the diagonal. By applying the reflection principle, it can be shown that the general form for the number of paths is 
$$\binom{2N}N-2\binom {2N}{m-2}+2\binom {2N}{m-n-3}-\cdots$$
where $\binom ab=0$ where $a<b$. 
However, for other degrees of overlap, it appears that the solution may be rather unwieldy. 

Question:
  Is there is a general form for the number paths for any degree of overlap i.e. a formula for the number of paths as a function of $n,m,q$, and if so, is there a closed form for such a formula?

 A: Not quite a closed form, but this seems to give a reasonable formula.
You have a series of diagonal "bottlenecks" each with $q+1$ vertices; each path uses exactly one vertex from each bottleneck. Let $M=(m_{ij})$ denote the $(q+1)\times(q+1)$ matrix where $m_{ij}$ counts the paths from the $i$th vertex in one bottleneck to the $j$th vertex in the next. Since these vertices are opposite vertices of a $(n-q+i-j)\times(n-q+j-i)$ rectangle, we have
$$m_{ij}=\binom{2(n-q)}{n-q+i-j}.$$
If there are $m\ge2$ chunks, there are $m-1$ bottlenecks. The product $M^{m-2}$ counts the paths from each vertex in the first bottleneck to each vertex in the last. To connect these bottlenecks up to the red dot points, we need to pre- and post- multiply by the vector of path counts connecting the start vertex to the first bottleneck, which is just the vector $v$ of binomial coefficients
$$\binom{2n-q}{n-i}$$ for $i=0,\ldots,q$. So the final number of paths is just
$$v M^{m-2} v^T.$$
In Mathematica:
M[n_,q_]:=Table[Binomial[2(n-q),n-q+i-j], {i,0,q},{j,0,q}];
f[n_,m_,q_]:=With[{v=Binomial[2n-q,n-#]& /@Range[0,q]}, 
v.MatrixPower[M[n,q],m-2].v]

For the example, we find $f(5,4,2)=22,036,472.$
The same method gives a formula if you have rectangles instead of squares, and can be used even if the overlap distances aren't constant as you go along.
