I think it's a little confusing because while every counting problem should have 1 numerical/number as an answer, there are sometimes different ways to get that answer that might seem different.

With the stars and bars trick, there are two ways (that give the same numerical answer after simplifying). You can have$$(i)\ \binom{\text{# of stars} + \text{# of bars}}{\text{# of stars}}\text{ OR } (ii)\ \binom{\text{# of stars} + \text{# of bars}}{\text{# of bars}}.$$

Anytime you answer a stars and bars question, either way works.

For example, in 5.2a the final numerical answer is 10. But you can get 10 with either (i) or (ii) above:$$(i)\ \binom{3+(3-1)}{3}=10 \text{ OR } (ii)\ \binom{3+(3-1)}{3-1}=10.$$

Similarly for 5.14, the final answer is 4, but$$(i)\ \binom{3+(2-1)}{3}=4 \text{ OR } (ii)\ \binom{3+(2-1)}{2-1}=4.$$

To review the other differences, 5.2a is 3 balls (so 3 stars) and 3 boxes (so 3-1 bars) while 5.14 is 3 balls (so 3 stars) and 2 boxes (so 2-1 bars). In both cases the number of bars is 1 less than the number of boxes.

Stars and bars is definitely tricky, so don't worry if it takes a little time to get use to it! Hope this helps!