In question is:
The number of possible ways a vehicle with capacity $K$ can pickup and drop off $n$ passengers in a single tour, while each passenger has an individual pickup and dropoff location.
Combinatorically, this seems to relate to both A000680 and A080934.
A000680 has the very simple and quite intutive formula $\frac{(2n)!}{2^{n}}$. This would be the number of valid entrance and exit sequences in a vehicle with unlimited capacity (ruling out all exit before entrance combinations). Alternatively, it could be modeled as the number of lattice walks in an n-dimensional lattice $(\mathbb{Z_{2}})^n$.
A080934 on the other hand limits the height of a dyck path. Also, for sequences of balanced parenthesss, this would mean the number of parentheses sequences of maximum depth $k$. For instance, (()()()) has depth 2, while (((()))) has depth 4. So, in the lattice example, it would be the number of subdiagonal walks in a two-dimensional lattice with distance from root location bounded by $k$. A recursive definition of this is explained in detail here. It is not completely congruent to my problem, since the bracket pairs are not distinguishable here.
Combining these together while staying in the lattice context, my formula should also be a calculation rule for the number of possible walks in $(\mathbb{Z_{2}})^n$ where the manhattan/hamming distance from the root location is bounded by $K$.
I even computationally filtered valid sequences for different values of $k$ and $n$ and searched OEIS for hours but only found sequences with small overlaps. Here is what I got for $k$, $n$ in the value range [0..5]:
$ \begin{array}{c|cccccc} k\backslash n& 0 & 1 & 2 & 3 & 4 & 5\\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 2 & 6 & 24 & 120\\ 2 & 1 & 1 & 6 & 54 & 648 & 9720\\ 3 & 1 & 1 & 6 & 90 & 1944 & 52920\\ 4 & 1 & 1 & 6 & 90 & 2520 & 99000\\ 5 & 1 & 1 & 6 & 90 & 2520 & 113400\\ \end{array} $
The diagonal matches $\frac{(2n)!}{2^{n}}$ (A000680), as the capacity is not a limiting factor here. The first row is simply $n!$, since only one can be in the vehicle.
I can give a method of calculation: consider
$$f_k(c,n)=n\,f_k(c+1,n-1)+c\, f_k(c-1,n)$$
but with $f_k(c,n)=0$ if $n<0$ or $c<0$ or $c>k$ and starting with $f_k(0,0)=1$,
designed to have $f_k(c,n)$ representing the number of ways of handling $c$ passengers already in the vehicle of capacity $k$ with $n$ people waiting to be transported, meaning that that $f_k(0,n)$ gives the values you want.
For example, with $k=4$, you get the following table for $f_4(c,n)$
$ \begin{array}{c|ccccccccc} c \backslash n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 0 & 1 & 1 & 6 & 90 & 2520 & 99000 & 4806000 & 274050000 & 17886960000 \\ 1 & 1 & 3 & 30 & 630 & 19800 & 801000 & 39150000 & 2235870000 & 145983600000 \\ 2 & 2 & 12 & 180 & 4320 & 140400 & 5724000 & 280260000 & 16012080000 & 1045548000000 \\ 3 & 6 & 60 & 1020 & 25200 & 824400 & 33660000 & 1648620000 & 94197600000 & 6150967200000 \\ 4 & 24 & 240 & 4080 & 100800 & 3297600 & 134640000 & 6594480000 & 376790400000 & 24603868800000 \\ \end{array} $
where $f_4(2,4)=4\,f_4(2+1,4-1) +2 \,f_4(2-1,4) = 4\times25200 +2 \times 19800 = 140400$.
So using the resulting $f_k(0,n)$ you can easily expand your table, starting with
$ \begin{array}{c|ccccccccc} k \backslash n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 2 & 6 & 24 & 120 & 720 & 5040 & 40320 \\ 2 & 1 & 1 & 6 & 54 & 648 & 9720 & 174960 & 3674160 & 88179840 \\ 3 & 1 & 1 & 6 & 90 & 1944 & 52920 & 1730160 & 65998800 & 2877275520 \\ 4 & 1 & 1 & 6 & 90 & 2520 & 99000 & 4806000 & 274050000 & 17886960000 \\ 5 & 1 & 1 & 6 & 90 & 2520 & 113400 & 6966000 & 525042000 & 46022256000 \\ 6 & 1 & 1 & 6 & 90 & 2520 & 113400 & 7484400 & 655678800 & 70146518400 \\ 7 & 1 & 1 & 6 & 90 & 2520 & 113400 & 7484400 & 681080400 & 80103945600 \\ 8 & 1 & 1 & 6 & 90 & 2520 & 113400 & 7484400 & 681080400 & 81729648000 \\ \end{array} $