Just want to put in a partial solution to provide an independent count to validate other solutions.
Note that I take both zero & dummy to be legal units.
A couple of general points:
If a (p,q)rider is present, then (p,q)leaper would be redundant.
If q>3, then a (p,q)rider can never do better than a (p,q)leaper, so stick with the latter.
Now consider the (0,1) direction.
With no riders, there are 2^7 = 128 possibilities.
If (0,1)rider is the only rider. There are apparently 2^6 = 64 choices of leaper but if all 6 leaper powers are present then (0,1)rider can never be blocked. So 63.
The only value of adding a (0,1)rider is to reach those locations that other riders and leapers cannot reach. We can apply this to other cases.
(0,2)rider and no other riders. Can’t have both (0,4)leaper & (0,6)leaper else (0,2)rider can never be blocked. So there are 3*2^4 = 48 choices of leapers.
Hence (0,1)rider & (0,2)rider gives 3*7 = 21.
(0,3)rider only is even simpler. Can’t have (0,6)leaper else (0,3)rider can never be blocked. So 2^5 = 32 choice of leaper.
Hence (0,1)rider & (0,3)rider gives 15 choices of leaper.
This leaves the corner case where (0,2)rider & (0,3)rider are combined. This offers unique feature of two routes from e.g. a1 to b7, so (0,6)leaper cannot be present. So 2^4 = 16 choices of leaper.
Hence (0,1)rider, (0,2)rider & (0,3) rider gives 2*3 = 6.
So I make it 128 + 63 + 48 + 21 + 32 + 15 + 16 + 6 = 329 possible combinations for the (0,1) direction. Please confirm.
The (1,1) direction will work similarly.