Bellman-Ford with Non-edge-decomposable Path Weights
No, Bellman-Ford won't work because the problem you described is NP-hard.

This is pretty easy to prove. I've been able to come up with several reductions using the same strategy. The general idea is to make it so that in the path you have several sections that can be chosen independently; then the overall question becomes whether you can coordinate all those choices so as to minimize the sum of the weights of the items in the chosen subpaths. You can easily set up the independent choices by using a sequence of "bottleneck" vertices $S = v_0, v_1, \ldots, v_{n-1}, v_n=T$ and making several paths between each $v_i$ and $v_{i+1}$ (the choices of which path piece from $v_i$ to $v_{i+1}$ to take for each $i$ are the independent choices I was talking about earlier).

For example, you can reduce from 3SAT. For each variable $x$, make one of the subpath choices be between item $x$ and item $\neg x$. For each clause, make one of the choices be between the items corresponding to the literals in the clause. For each variable either the variable, the negation, or both must be in any path. The only way to get away with having only one version of each variable in the path is if there exists an assignment that makes every clause true. If each item has weight 1, then the overall weight of the items in the path will equal the number of variables iff the 3SAT instance is satisfiable.

As another example, you can reduce from set cover. For each set that you want to cover, make a piece of the path be a choice that lets you chose any of the items in that set. Then to minimize the number items chosen in all, you have to cover all the sets with as few items as possible.