The top-down columns mark the same point in time, the left-right lines/rows the same variable/sub formula. The bigger block in this example is said to be repeated forever.

When evaluating a subformula at a position in the flow diagram, that means that we take this place (this x-axis) as “now” (t_{0}).

About q1 ([a SU b]) you are right. We only have 1 from which on b follows after finite time, and a is present on the way there. Since we don't have b at all, q1 remains false.

About q2 ([q1 WU c]) you are not right. It is correct that q2 will become 1 where c is true, also it would become true on the way to where c is true (if q1 holds all the time on the way there), and (different from SU) q1 also holds where G q1 holds (place from which q1 is true forever). Thus, [q1 WU c] is only true in the first step. After that it is not, because we never see c again.

About q3 ([q2 SU d]) you are only partly right. It is correct that in the third step and on the way to the third step, the formula would be evaluated to 1 if we had d in the third step, and q2 on the way to it. However, from the fourth step on, the formula evaluates to false as d is not seen from there.

Where is your mistake? I think the mistake is that you do not isolate the time steps. As I said before, for every time step in the diagram, we consider this point as “now” and don't look back to the past for future operators. See also slide 9 in the exercise slides on exercise 8 (temporal logic).