I haven't seen any attempts to do so, probably because there would be 64 entries in the table. Another reason they don't show up much is because we don't usually care much for uniqueness or totality in relation counting, we usually care more about stuff like symmetry, anti-symmetry, and transitivity. But, as I have nothing better to do, I'll give it a go here!
So the first thing to notice that we can switch all the rights with lefts, and the formulas should hold, so we in fact only need 48 cases.
I'll sort these into $3$ different $4\times4$ tables, one for each distinguishable/indistinguishable combo. This gives even more symmetry in the all distinguishable/indistinguishable tables so we actually only require $36$ entries. Thus, a catchy title for this might be "The $36$-fold Way"
Time to count relations between nonempty sets $|X|=n$ and $|Y|=m$ (subsets of $X\times Y$).
First table, everything is distinguishable, and $S(n,m)$ are the Stirling Numbers of the second kind:
$$\begin{array}{c|cccc}
\text{left}^{\large{\text{right}}} & \text{none} & \text{unique} & \text{total} & \text{both} \\
\hline
\text{none} & 2^{nm} & \sum_{k=0}^n \binom{n}{k}m^k=(m+1)^n& (2^{n}-1)^m& n^m \\
\text{unique} &-& \sum_{k=0}^{\min(n,m)} \binom{m}{k}\binom{n}{k}k!&\sum_{k=m}^n \binom{n}{k}m!S(k,m)
& m!\binom{n}{m} \\
\text{total} &-&-&\sum_{k=1}^{n}(-1)^{n-k}\binom{n}{k}(2^k-1)^m& n!S(m,n)\\
\text{both} &-&-&-& \begin{cases}n! & n=m\\0 & n\neq m \end{cases} \\
\end{array}$$
An alternative for the $n!S(m,n)$'s: we could use PIE to count this as $\sum_{k=1}^n (-1)^{n-k}\binom{n}{k}k^m$, but this would require a double summation, which I'd rather not have in my tables if I can help it.
One application of this could be counting the number of ways $n$ people can participate in $m$ clubs. Then right totality for example would mean each club has at least one member.
Warning: Past this point, stuff gets weird, as clubs or people become indistinguishable. It also gets hard to count, as there are many different symmetries which can make things indistinguishable. Thus, I have not filled out most of these others currently, but I'll try to update these over time, and would appreciate help in the comments!:
Left is distinguishable, right is not:
$$\begin{array}{c|cccc}
\text{left}^{\large{\text{right}}}& \text{none} & \text{unique} & \text{total} & \text{both} \\
\hline
\text{none}&&\sum_{k=0}^m \binom{k+n-1}{k}&&\binom{m+n-1}{m}\\
\text{unique}&&\sum_{k=0}^m\binom{n}{k}&&\binom{n}{m}\\
\text{total}&&\sum_{k=n}^m\binom{k-1}{k-n}&&\binom{m-1}{m-n} \\
\text{both}&
\sum_{k=0}^{m}S(n,k)&
\begin{cases} 1 & n\leq m\\ 0 & n < m \end{cases}&
S(n,m)&
\begin{cases} 1 & n=m \\ 0 & n\neq m \end{cases}
\end{array}$$
And finally, everything is indistinguishable:
$$\begin{array}{c|cccc}
\text{left}^{\large{\text{right}}}& \text{none} & \text{unique} & \text{total} & \text{both} \\
\hline
\text{none}&&\sum_{k=0}^n \pi(k,m)&&\sum_{k=0}^m \pi(n-k,k)\\
\text{unique}&-&\min(n,m)+1&\pi(m,n)&\begin{cases} 1 & n\geq m \\ 0 & n < m \end{cases}\\
\text{total}&-&-&&\pi(n-m,m)\\
\text{both}&-&-&-&\begin{cases} 1 & n=m \\ 0 & n\neq m \end{cases}
\end{array}$$
Where $\pi(m,n)$ is the number of ways to partition $m$ into at $n$ (possibly empty!) parts.
