In the same lecture notes without providing many details it says that the complexity of the algorithm which uses a balanced search tree is $O(n\log n+R)$ where $R$ is the total amount of intersections.
However I don't understand why this is the case.
Suppose that the sweep line is travelling from top to bottom. Whenever we encounter a vertical line segment, we pick its $x$ coordinate and add it to a search tree. Assuming that we are using a balanced search tree like AVL, this operation will take $O(\log n)$ time. Now when we reach an end point of a vertical line segment, we remove its $x$ coordinate from the search tree, which is going to take $O(\log n)$ time as well.
Whenever we encounter a horizontal line segment defined by points $A$ and $B$, we have to do a range search on our search tree and use the range defined by these two points, that is $(A.x, B.x)$
Now how is this range search going to happen? We use $A.x$ and travel to the left most point in the search tree, and do the same for $B.x$, we can do this in $O(2\log n)$.
All the points in our tree between these two end points are going to be the intersecting the horizontal segment, which is going to be let's say $R$ in total, so $O(2\log n+R)$ in total.
We do this $n$ times (the amount of horizontal segments) so we get a complexity of $O(2n\log n+nR)$.