Why it is important the notion of equivalent divisors in pairing definitions?

Following the book Pairing for Beginners, the Tate pairing computation requirements are:

1. Let $P$ be an point on the $r$-torsion subgroup in $E(\mathbb{F}_q)$.
2. Let $f$ be a function whose divisor is $(f) = f(P) - r(\mathbb{O})$.
3. Let $Q$ be a point of $E(\mathbb{F}_{q^k})$.
4. Let $D_Q$ be a degree zero divisor that is equivalent to $(Q) - (\mathbb{O})$, with disjoint support to the one of $(f)$.

The pairing is then defined as $e(P,Q) = f(D_Q)$.

My questions are about the 4-th requirement: Why is the notion of equivalence important here? Why could not fix the divisor $D_Q$ to be, say, $D_Q = ([2]Q) - (Q)$? Should not the pairing value change if we evaluate $f$ on another (correct) divisor $D_Q'$ while mantaining the point Q? Or is it that $f$ will output the same once evaluated on two divisors $D_Q,D_Q'$ that are both equivalent to $(Q) - (\mathbb{O})$ (i.e., they belong to the same class in the Picard group $Pic^0(E) = Div^0(E)/Prin(E)$)?