Let's say, we initially had some positive numbers $a$ and $b$ such that:
$$\frac{a}{b}=r$$
Now $a$ and $b$ have changed by a certain percentage, meaning:
$$a_1=a \cdot (1+x)$$
$$b_1=b \cdot (1+y)$$
Where $x$ and $y$ are real numbers (the percentages are given by $100 \cdot x$ and $100 \cdot y$), which could be negative or positive, corresponding to decrease or increase. We can take them to be $ \in(-1,1)$ so the change is less than $100$ %.
Then we have:
$$\frac{a_1}{b_1}=\frac{a \cdot (1+x)}{b \cdot (1+y)}= \frac{1+x}{1+y} \cdot r$$
Speaking about percentage again, write the change in $r$ as:
$$r_1=r \cdot (1+z)$$
Where:
$$z=\frac{1+x}{1+y}-1=\frac{x-y}{1+y}$$
Your question is "what contributed more to the change in ratio?" It is not really clear to me what you are asking. But there are ways to see how $x$ and $y$ contribute to the change separately.
Let's consider your case, initially:
$$x=0 \\ y=0$$
And:
$$\Delta x=-0.2 \\ \Delta y=0.2$$
Then, applying the changes separately we have:
$$\Delta_x z=\frac{x+\Delta x-y}{1+y}-\frac{x-y}{1+y}=\frac{0-0.2-0}{1+0}-\frac{0-0}{1+0}=-0.2$$
$$\Delta_y z=\frac{x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2}{1+0.2}=-0.16666\dots$$
It seems that the change in $x$ contributed more in this case. But you can try some other cases to see what happens.
Applying the changes at the same time we have:
$$\Delta z=\frac{x+\Delta x-y-\Delta y}{1+y+\Delta y}-\frac{x-y}{1+y}=\frac{-0.2-0.2}{1+0.2}=-\frac{0.4}{1.2}=-\frac{1}{3}=-0.33333$$
The change in ratio was actually $ \approx -33.3 $%, not $-50$ % as you seem to believe.
Best Answer
For the exact change in the ratio we have, $$\Delta\left(\frac ab\right) = {a+\Delta a \over b+\Delta b} - \frac ab = {b \Delta a - a \Delta b \over b(b-\Delta b)}.$$ Although you can get a term that only contains $\Delta b$ from this, there’s no way that I can see to get one that involves only $\Delta a$.
For small values of $\Delta a$ and $\Delta b$, the change in the ratio can be approximated by the differential: $$D\left(\frac ab\right)[\Delta a,\Delta b] = \left(\frac1b,-\frac a{b^2}\right)\cdot\left(\Delta a,\Delta b\right) = {b\Delta a-a\Delta b\over b^2}.$$ For your example, this approximation predicts a change in the ratio of $0.625$, almost double the actual change, so is unfortunately not likely to be of use to you.
A possible way to separate the effects of $\Delta a$ and $\Delta b$ is to look at the change in the logarithm of the ratio instead: $$\ln\left({a+\Delta a \over b+\Delta b}\right)-\ln\left(\frac ab\right) = \ln\left(1+\frac{\Delta a}a\right)-\ln\left(1+\frac{\Delta b}b\right).$$