You are wondering about the successor state of ({s3},{}) for input {}. For the first component, we compute the successor states for {s3} under input {} which is {s0}. The computation of the second component depends on the value of the second state set in ({s3},{}). Since it is empty, the second component is computed as intersection of our first component with the accepting states, hence, we compute {s0} ∩ {s0,s1} = {s0}. Hence, we have the transition from ({s3},{}) to ({s0},{s0}) under input {}.
So, you are not right. We just compute compute succ({s3}) as first component and succ({s3}) ∩ F as the second component. That's all.
But I saw that there is a but in the deterministic automaton: It is impossible that there are two transitions leaving state ({s3},{}) for the same input {}. The transition from ({s3},{}) to ({s2},{}) is taken for input {a}. The example solutions have been updated.