Solve the Trolley Problem

Python 3, 80 bytes

y=lambda d,s=0,p=[],f=0:f in p and s or min(y(d,s+d[f][t],p+[f],t)for t in d[f])

Try it online!

Takes input as a dictionary keyed by node id. The entries are a dictionary of neighbors and the number of people on the track between a node and the neighbor. E.g., for the first test case:

{0: {1: 0}, 1: {2: 5, 3: 1}, 2: {2: 0}, 3: {3: 0}}

0 is the start node, 1 is node 'a', 2 is node 'b', etc.

Jelly, 27 23 bytes

ṗL$µṭ0FIm2ASµÐḟµQ⁴ySµ€Ṃ

Try it online! (last test case)

Cruel Version (Mutilate the most people)

Takes input as numbers. For the last example, 1 is a and 2 is b. 0 is the starting node. The first argument is the list of edges (e.g. [0,1],[0,2],[1,1],[2,2]), and the second argument is a list of the edges and the number of people on them (e.g.[[0,1],[0,2],[1,1],[2,2]],[1,1,0,0]).

How it Works

ṗL$µṭ0FIm2ASµÐḟµQ⁴ySµ€Ṃ
ṗL$                       - on the first argument, get the Cartesian power of it to its length.
                            this gives all paths of the length of the input. Cycles are implicit
   µ        µÐḟ           - get valid paths starting with 0 -- filter by:
    ṭ0                      - prepend 0
      F                     - flatten
       I                    - get the difference between elements
        m2                  - every second difference: 0 for each edge that starts at the node the previous edge ended at, nonzero otherwise.
          AS                - 0 iff all elements are 0
               µ    µ€    - on each path:
                Q           - remove repeat edges.
                 ⁴y         - translate according to the mapping in the second program argument
                   S        - Sum
                      Ṃ   - get the minimum of these.