By contradiction, we can prove that for a connected undirected graph, each edge belongs to at most one cycle if and only if there are at most two non-intersect paths between any two vertices. In addition, the maximum flow problem and the minimum cut problem are equivalent, so we only need to consider how to cut some edges with the lowest sum of capacity so that the source vertex $$$s$$$ is not connected to the sink vertex $$$t$$$.
Let's take a look at the $$$s$$$-$$$t$$$ cut problem. If there are at least two edges in one cycle that are cut, we could figure out by adjusting that at most two of them should be cut, since there are at most two non-intersect paths between $$$s$$$ to $$$t$$$. In addition, if there are many cycles where some edges of each of them are cut, it can be known by adjusting and contradiction that only the edges of at most one cycle of them should be cut.
One observation is that if there is at least one edge in some cycle should be cut in the best solution of the minimum cut problem, there must be two edges in that cycle need to be cut, one of which is the edge with the smallest capacity in that cycle. Hence, we can remove the edge with the smallest capacity in each cycle and add its capacity to the other edges in the cycle, and the maximum flow between any two vertices will not change in value.
The modified graph forms a tree. If we add edges to an empty graph in descending order of its capacity, we can determine $$$\mathrm{flow}(s, t) = w$$$ $$$(s \in S, t \in T)$$$ for the two sets of vertices $$$S$$$ and $$$T$$$ that are firstly connected after joining the edge of capacity $$$w$$$. Just maintain these sets of vertices by the disjoint set union, count in each set the number of vertices whose bit in some fixed position are $$$1$$$, and then it is easy to calculate the answer by enumerating each bit. Time complexity is $$$\mathcal{O}(m \log n)$$$.