There are several problems in your recurrence:
for $h = 0$ or $w = 0$, there should be one possibility, not zero: the only way to build a tower of height or width zero is adding no block at all;
you write:
Find the number of towers of height 'h - 1' and of the same width. That function is Fun(h - 1, w). The last height has the block size of 1x2, this can be filled in two ways, two 1x1 blocks and one 1x2 block.
but if you are considering block size 1×2, that means that you are implicitly considering $w = 2$, which is not necessarily the case;
you are not considering the possibility to add a (for example) $2\times 2$ block at the top of the tower, in the case $w = 2$.
What I would suggest:
- write a function to compute the number of possibilities for a tower of width $1$ and height $H$, independently (this is easier to write and can serve as a preliminary exercise). Let us denote this function $f_1(H)$;
- to build a tower of height $H$ and width $2$, for $H> 0$, you have to consider the top block of width $2$ that appears in the tower. If this block ends at height $h\geqslant 0$ and has height $k$, we have to complete with two towers of width $1$ and height $H - h - k$. That meanscases:
$$f_2(H) = \sum\limits_{h=0}^H\sum\limits_{k=0}^{h}f_2(h) \times f_1(H-h-k)^2$$
- if there are no block of width $2$, there are $f_1(H)^2$ possibilities to construct the tower (two towers of width $1$ and height $H$);
- if the top block of width $2$ that appears in the tower starts at height $h\geqslant 0$ and has height $k$, we have to complete with two towers of width $1$ and height $H - h - k$, over one block $k\times 2$ over a tower of height $h$ and width $2$. That means: $$f_2(H) = \sum\limits_{h=0}^{H-1}\sum\limits_{k=1}^{H-h}f_2(h) \times f_1(H-h-k)^2$$