I can try explaining Fixed-Length Paths 1. The pseudocode for it looks like this.

0 for every vertex u in the centroid tree:
1     map<int, int> count_dist
2     for every child v of u in the centroid tree:
3         for every descendant w of v in the centroid tree:
4             dist = distance between u and w in the original tree
5             ans += count_dist[K - dist]
6 
7         for every descendant w of v in the centroid tree:
8            dist = distance between u and w in the original tree
9            count_dist[dist] += 1

First, complexity is $$$O(n \log^2 n)$$$ (it may be possible to remove one of the logs). Why is that the complexity? For every vertex $$$u$$$ in the centroid tree, you go over all of its descendants in the centroid tree. Conversely, this takes the same amount of time as going over each ancestor of each vertex. Every vertex has at most $$$\log n$$$ ancestors, so this takes $$$O(n \log n)$$$ work at most. The second log comes from using map and computing distances, both of which you may be able to get rid of.

Why does this work? On line 5, you are essentially adding, for each $$$w$$$, the number of paths with length $$$k$$$, such that the LCA of $$$w$$$ and the vertex at the other end in the centroid tree is $$$u$$$. You want to avoid adding vertex pairs where the LCA is lower in the centroid tree; that's why you go over every child $$$v$$$ of $$$u$$$, as opposed to just iterating over all descendants of $$$u$$$.