When calculating a standardized change score, what do you divide by and why? I understand that, in most cases, one calculates a standardized change between two scores as

(Lakens 2013)
i.e., the mean of the change scores is divided by the standard deviation of the change scores.
My question is: Why don't we divide by the standard deviation of the scores at baseline, instead? It seems more intuitive to me to present a standardized change as a measure of variability in the scores at baseline rather than variability in how much scores changed.
For example, imagine a medication for headaches; at baseline, people's headache pain scores have a mean of 10 and a standard deviation of 2. After taking the medication, their headache pain scores have a mean of 8. The drug works about the same in everybody; the standard deviation of the change score is only .25. Wouldn't it be more useful to say "this drug helped move the average person a whole standard deviation down from their baseline pain score" rather than "this drug moved the average person down eight times what the average person changed"?
What is the rationale here?

A Cohen's $d$ is a ratio: the observed effect divided by a standard deviation (SD). The SD is a "unit of measurement" and you are correct that there are different choices for the SD in the denominator of a Cohen's $d$.
For example, the (Lakens, 2013) paper introduces three Cohen's $d$ effect sizes for within-subject comparisons (and generally recommends to report $d_{rm}$ or $d_{av}$, not $d_z$).
$$
\begin{aligned}
d_z
&= \frac{M_\text{diff}}{\sqrt{\operatorname{SD}_\text{pre}^2 + \operatorname{SD}_\text{post}^2 - 2r\operatorname{SD}_\text{pre}\operatorname{SD}_\text{post}}} \\
d_{rm}
&= \frac{M_\text{diff}}{\sqrt{\operatorname{SD}_\text{pre}^2 + \operatorname{SD}_\text{post}^2 - 2r\operatorname{SD}_\text{pre}\operatorname{SD}_\text{post}}}\sqrt{2(1-r)} \\
d_{av}
&= \frac{M_\text{diff}}{\left(\operatorname{SD}_\text{pre} + \operatorname{SD}_\text{post}\right) / 2}
\end{aligned}
$$
where $\operatorname{SD}_\text{pre}$ is the standard deviation at baseline (before treatment), $\operatorname{SD}_\text{post}$ is the standard deviation after treatment and $r$ is the correlation between the measures.
Your argue in favor of another variant:
$$
\begin{aligned}
d_{?}
&= \frac{M_\text{diff}}{\operatorname{SD}_\text{pre}}
\end{aligned}
$$
Note that if the treatment doesn't change the standard deviation, then $d_{?} \approx d_{av}$. However, it might be the case that the treatment is very effective for some patients while not effective at all for others. That is, the $\operatorname{SD}_\text{post} > \operatorname{SD}_\text{pre}$. The Cohen's $d$ you propose won't be sensitive to the increase in variability and $d_{?} > d_{av}$. (Also consider the case where the variance decreases post-treatment.) The question is: If that were the case, do you think $d_{?}$ or $d_{av}$ is a better effect size measure?
[1] D. Lakens. Calculating and reporting effect sizes to facilitate cumulative science: a practical primer for t-tests and ANOVAs. Frontiers in psychology, 4:863, 2013. 
