My guess is the difference in the number of intersections from one addition of a transversal to before that transversal was added is given by $f(n+1)-f(n)$
I conjecture $2\leq f(n+1)-f(n) \leq n$ Please correct me if I'm wrong.
You add n because at most, the number of intersections anyone can add is by that new transversal going thru all existing lines.
Zero lines: zero intersections.
One line: zero intersections
Two parallel lines zero intersections.
two parallel lines, add one transversal you have two intersections. Let that number be f(3)=2.
define a function f(n) where n is the number of current lines you have for $n \geq 3.$ and $f(n)$ is the number of total intersections of all lines.
Add the next transversal, so f(4)=f(3)+3.
So on, f(n+1)=f(n)+n for $n\geq 3.$
This is the upper bound.
On the lower bound, keep adding transversals parallel to the first transversal, you'll add two intersections every time so f(n+1)=f(n)+2 for $n \geq 3$.
I conjecture an intermediate case which fits between the bound of add two every time, or as in $f(n+1)=f(n)+n$ is when after the first transversal added, you added next transversal by going thru the same point of intersection with the parallel lines as one of the previous transversals. This intermediate case doesn't minimize because sometimes it added more intersections than all parallel line approaches, and I don't know whether it can beat the upper bound formula $f(n)+n=f(n+1)$. Please let me know how the intermediate case behaves, if there's some kind of pattern.