This is the second jump. We start at position B, where Z is now.
We compute Z2. We use the same technique as before. The first triangle is the one whose corners are A, 1 and B. we rotate this triangle around A to produce a new similar triangle whose corners are A, B and D. "Similar" means that it has the same proportions as the first triangle.
Then we add on the value of C, which is the line from D to E (same length and direction as the line from A to C).
We won't follow this particular sequence any further, but if you did then you would find that Z (each time through the loop) moves around, bobbing in towards the centre and out again, but never crossing the boundary of the map (we'll think about why it doesn't shortly).
Because Z never crosses the boundary of the map, the original position C (not Z) is deemed to be part of the Mandelbrot Set. And if you check here, you will see that it is.