AF AG a is best understood by decomposing into two problems: First compute the set of states phi=AGa which is only state s3. Now, you can consider AF s3, i.e., the states where all outgoing infinite paths eventually reach s3. The state s0 does not satisfy this, since you can loop infinitely often in s1, and for the same reason, it does not hold in s1. AF s3 only holds in states s2 and s3.

Assume the Kripke structure would be labeled such that only state s3 would be labeled with a variable p, and you would consider AF p. Would you then also think that states s0 and s1 would satisfy that formula?