Number of Regions in the Plane defined by $n$ Zig-Zag Lines

$ZZ_n = ZZ_{n-1} + n + 8(n-1)$.

Add one new straight line first, intersecting the previous $n-1$ lines. This gives the term $n$. Then make the small ziz-zags as suggested in the hint. This adds an additional $8$ regions at each of the $n-1$ intersections.

Your observation is correct. If you believe the first step, then there are $9(n-1)$ intersections of the new zigzag with the existing constelation. Now any two consecutive intersection points cut an existing bounded region into two regions, resulting in a total of $9(n-1)-1$ regions. Similarly the line segment between the first intersection and infinity and the line segment bewteen the last intersection and infinity each cut an existing unbounded region into two. So there are $9(n-1)-1+2$ new regions.

What also works is:

When you assume first assume that, for every zig zag line, there are two parallel infinite lines and one infinite line intersecting both (so we extend our line segments).

Now, for the case of the zig zag lines, every parallel line stops at the intersection with the third line, so you have to subtract 2 regions from $L_{3n}$ for every zig zag line. The third line is bounded in two directions, so you have to subtract two additional regions for each zig zag line. The last step is to subtract 1 region because the two parallel lines don't intersect.

So $$ZZ_n = L_{3n} - 5n = \frac{9n^2+3n}{2}-5n+1 = \frac{9n^2+3n}{2} - \frac{10n}{2} +1 = \frac{9n^2-7n}{2} + 1$$