The adversary $A$ breaks scheme $\Pi$ by generating some kind of forgery.
Every forgery can be assigned some label $i$.
This label $i$ is chosen by the adversary, but there are only a polynomial number of choices for $i$.
The reduction algorithm can set things up to look just like the world that $A$ expects.
Additionally, the reduction algorithm can set things up with a specific $i^*$ in mind, so that if the adversary produces a forgery whose label happens to be $i^*$, then the reduction algorithm can successfully break scheme $\Theta$.
Importantly, and this may be the thing you're missing, $A$'s view of things is independent of $i^*$.
In other words, no matter what specific $i^*$ the reduction has in mind, it always produces a perfectly faithful simulation of the world that $A$ expects.
How should the reduction algorithm choose $i^*$?
It cannot predict in advance what label the adversary's forgery will have.
So instead it will choose $i^*$ uniformly at random from among the polynomially many choices.
Why does this work?
Eventually the adversary outputs a forgery, and the forgery has some label $i$.
If the adversary's entire view is independent of the choice of $i^*$, then the adversary's choice of $i$ is independent of the choice of $i^*$.
Since $i^*$ is uniformly distributed and independent of $i$, we can say that $\Pr[i=i^*] = \frac{1}{\mbox{# of choices}}$.
In your setting, the "label" $i$ of a forgery is (apparently -- I followed your instruction to not actually read the paper) the index of the first signing-oracle query that is made using the time period named in the forgery.
If the reduction algorithm can predict which signing-oracle query will be the first one made under the time period of the adversary's forgery, then it will do something special in response to that query (embed some information that will help it break $\Theta$).
Of course it cannot predict this property of the forgery, so it guesses.
There are only $q(k)$ possibilities for the identity of this "special" query.
In your setting, there is also some aborting going on, but this is a distraction to the real probability question.
What's really going on is this:
The reduction is only successful in breaking $\Theta$ when its guess $i^*$ equals the label $i$ of $A$'s forgery.
The authors here are saying that the reduction might as well give up (i.e., abort) when it sees that its guess is going to be wrong.
It would have been fine if they wrote the reduction algorithm to never abort, and instead just make a blind "stab in the dark" at breaking $\Theta$ in these cases.
