Note: the word compute should be emphasized here because the purpose of this question is to request an algorithm that I can implement in C. My current algorithm generates almost all of the weight assignments. My goal is to generate all of them (for arbitrary $m$ and $n,$ up to the limits of my hardware).
Let $P$ be a poset on $n$ points. It can be assumed throughout that we have the automorphism group $\text{Aut}(P)$ of $P.$ Let the orbits under $\text{Aut}(P)$ have lengths $a_1,a_2,\dots,a_k.$
Let $C_{i,j}$ and $P_{i,j}$ be the sets of compositions and partitions, respectively, of $i$ with $j$ nonnegative parts.
One might guess that by assigning all compositions in $C_{m,k}$ to the set of orbits (one part per orbit) and to each such composition $c_1+c_2+\dots+c_k,$ all possible combinations of partitions $\pi_{i,1}+\dots+\pi_{i,a_i}\in P_{c_i,a_i}$ (one part per vertex) for $1\leq i\leq k,$ all equivalence classes under $\text{Aut}(P)$ of vertex weight assignments that sum to $m$ would be obtained. It turns out that for small $m$ and $n$ they would be right: the above algorithm produces all weight assignments for $m+n\leq5$.
However, when $n=4$ and $m=2,$ it misses exactly one. Since $\text{Aut}(P)=\{(1), (0\ 1)(2\ 3)\}$ for the poset $P$ in the diagram below, its orbits are $\{0,1\}$ and $\{2,3\};$ our algorithm fails to recognize it matters whether we select vertex $2$ or $3$ to go with vertex $0$ (assuming it was the one selected from $\{0,1\}$ under the composition $1+1$ of $m$ and its only possible combination of subpartitions, namely, $1+0$ for each orbit).
As would be expected, this theme gets repeated for $m=2$ as we increase $n$:
What's the algorithm missing? What does it additionally need to generate all of the weight assignments?
==== info below added 18 May 2021 ====
A year and a half ago I posted a message to Brendan McKay's nauty mailing list asking whether comprehensive lists of finite topologies on more than 7 points were available anywhere. Our correspondence can be viewed here. As per his reply, McKay generously sent me files containing all transitive digraphs on up to 9 points, from which I was able to easily (using his detailed and easy-to-follow instructions) produce comprehensive lists of all nonhomeomorphic finite topologies on up to $9$ points (available here).
Unfortunately, McKay did not provide any details about the method he used to compute the digraphs and I didn't ask. His and Brinkmann's paper Counting Unlabelled Topologies and Transitive Relations shows how to enumerate them in terms of vertex-weighted posets, not how to compute them.
His files evidence that they can be computed easily though, somehow. I "reverse engineered" them to arrive at the algorithm above. It's interesting how few weight assignments it misses: for $m+n=6$ just $1$ out of $400,$ for $m+n=7$ just $7$ out of $2490,$ and for $m+n=8$ just $57$ out of $18980.$
Most if not all of the missing ones in these cases are easily derivable from the Hasse diagrams above. For example, consider the middle poset in the second row. Since the only troublemakers are $2\rightarrow4$ and $3\rightarrow5$ we can remove all combinations of the four pairs of edges $\{0\rightarrow4,0\rightarrow5\}$, $\{1\rightarrow4,1\rightarrow5\}$, $\{0\rightarrow2,0\rightarrow3\}$, $\{1\rightarrow2,1\rightarrow3\}$ to get $16$ of the $57$ missing weight assignments. The same can be done with its dual, though this does not produce $16$ more since for example removing all four pairs produces the same thing in both cases. Similar removals from its self-dual neighbor to the right account for still more (but again fewer than $16$).
==== info below added 19 May 2021 ====
Just cross-posted on CSSE here.