You are right to ask these questions; we usually have more time for the DFS recitation, and I would definitely have liked to go over that part in more detail. I will indeed go over the algorithm in more detail, in the second DFS recitation.
I will answer your questions here as well. Actually they are all related, so I will explain the whole process after the topological sorting. Please open the presentation to page 14, as I will refer to it.
(Actually we only pass each edge at most once - I will update the slides)
We want to find out if there is a path, however we clearly can't go over all the possible paths, because there may be exponentially many paths.
So we topologically sort, and then it suffices to check whether there is a path from each SCC (strongly connected component) to the next (make sure you understand why!)
To do this, we use BFS because it reaches the vertices in the order of their distance from the source. Take a look at the BFS from A. We need to check if there is a path between A and D. We reach B and F first. Now we notice that F is after D in the topological sort, so we don't advance further in that sub tree. We continue from B, reach C and G, notice G is not in the subgraph betwen A and D and stop exploring that sub-tree. Ten we continue from C to D.
Now we run a BFS in the sub tree between D and F, and so on.
Notice that we check each edge at most once. We say that we check each vertex at most twice, because it can appear in at most two sub-graphs. (Notice that when we reach F from A, we don't count it as having reached F, because that would complicate our analysis. But we don't have to, because that operation took O(1) - crossing the edge takes O(1), and we assume that we can compare two vertices to find which comes earlier in the sort in O(1).)
I hope that answers your questions. I am not sure exactly what you meant by Question 3. I think I answered it, but if not, please ask again and word it slightly differently, because I am not sure whether I understand the question in its current wording.