In a Boolean function $f$, it seems that the number of terms in a Disjunctive Normal Form (DNF) is equal to the number of truth outputs of $f$. For example,
If $f=A+B$, then clearly, $f$ has $2^2-1=3$ truth outputs.
Also, DNF of $f$ is $f=AB+A\bar{B}+\bar{A}B$ (three terms). Likewise, we can show that DNF of $f=A+B+C$ contains $7$ terms, but its number of truth outputs is also $7$.
Is it true in general that the number of terms in a DNF is equal to the number of truth outputs of $f$?
If we are talking about DNF in general that's not true. For instance, $(A\vee\neg A)\vee B_1\vee\dots\vee B_n$ is identically true but one of it's DNF is $A\vee\neg A$. Also you can duplicate conjunctions: $$A\vee A\vee\dots\vee A\vee\neg A\vee\neg A\dots\vee\neg A$$
and disjuntions:
$$(A\vee\neg A)\vee(A\vee\neg A)\vee\dots\vee(A\vee\neg A).$$
But if we are talking about perfect DNF then yes, you are right. In perfect DNF each conjunction includes all the variables of the formula and there are no duplicates. The reason is the following: we build a perfect DNF looking at a truth table and matching a unique conjunction to each of the truth outputs.