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:
Take 0 =< p =< q
If a (p,q)-rider is present, then (p,q)-leaper would be redundant.
If q > 3, then a (p,q)-rider could never do more than 1 leap, instead use (p,q)-leaper.
Just going to look at the hardest case where 7 leaps are possible, e.g. the (0,1) direction.
With no riders, there are 2^7 = 128 possible sets of leapers. The empty set doesn’t allow movement at all in this axis: that’s ok. Maybe the movement is in other directions or if not we have the immobile dummy.
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.
Note that 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 pattern to other cases. Instead of a * 2^b combinations we have when we add the (0,1)-rider: a * (2^(b-1)-1).
(0,2)-rider and no other riders. Can’t have both (0,4)-leaper & (0,6)-leaper else (0,2)-rider power is never needed, and might as well have (0,2)-leaper. 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-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.