Let $E,N$ and $D$ denote the number of moves in East, North and diagonal direction respectively. Then a solution must satisfy
Note that for specific $(i,j)$, $D$ determines the system entirely. So given $D$ we can calculate $E,N$.
Now, for a given tuple $(E,N,D)$ satisfying the above for a given endpoint $(i,j)$, it becomes a combinatorial question how many different orders we can form using these letters. In the case of $(i,j)=(2,2)$ and $(E,N,D)=(1,1,1)$ we have the possibilities
DNE (path 1)
DEN (path 2)
NDE (path 3)
so of course that makes $3!=6$ paths. Here path 1 and 2 correspond to your first diagram and path 3 to your third. Indeed path 3 is the only path with the diagonal connecting $(0,1)$ and $(1,2)$, but just regarding this as permutations of letters we do not need to be that concerned with specific locations of specific parts of the paths.
I think the formula
could be relevant here. Note that $(i-D)=E$ and $(j-D)=N$ which is why those figures appear. So for a given $D$ this should calculate the number of paths that can be formed.