As mentioned above, I am trying to take the first-order Taylor expansion of $$\frac{f(x_1+dx_1)}{f(x_2+dx_2)}$$ My attempt is
$$\frac{f(x_1) + \frac{df}{dx_1}\Delta x_1}{f(x_2) + \frac{df}{dx_2}\Delta x_2}$$ where I've just taken 'separate' Taylor expansions on the numerator and denominator. I don't think this is right? It'd be great if someone could point me in the right direction, thanks!
EDIT - POSSIBLE ANSWER:
This seems like a plausible attempt:
$$ [f(x_1 + dx_1)][f(x_2 + dx_2)]^{-1} $$
Taking Taylor expansions, we have
$$ \left[f(x_1) + \frac{df}{dx_1} \Delta x_1 \right]\left[ f(x_2) + \frac{df}{dx_2}\Delta x_2 \right]^{-1} $$
$$ \left[f(x_1) + \frac{df}{dx_1} \Delta x_1 \right] \frac{1}{f(x_2)} \left[ 1 + \frac{1}{f(x_2)} \frac{df}{dx_2}\Delta x_2 \right]^{-1} $$
Then taking the binomial expansion for the 3rd term in the product, we get
$$\frac{1}{f(x_2)} \left[ f(x_1) + \frac{df}{dx_1} \Delta x_1 \right] \left[ 1 - \frac{1}{f(x_2)} \frac{df}{dx_2} \Delta x_2 \right]$$
We can then expand and get rid of the $\Delta x_1\Delta x_2$ term because they are small terms. I eventually get
$$\frac{f(x_1) - \frac{f(x_1)}{f(x_2)}\frac{df}{dx_2}\Delta x_2 + \frac{df}{dx_1}\Delta x_1}{f(x_2)}$$
For (my) convenience, I changed notation and, in the same spirit as in your edit, I considered $$\frac{f(x+a)}{f(y+b)}$$ and used Taylor expansions around $a=0$ and $b=0$.
So, $$f(x+a)=f(x)+a f'(x)+O\left(a^2\right)$$ $$f(y+b)=f(y)+b f'(y)+O\left(b^2\right)$$ $$\frac 1{f(y+b)}=\frac{1}{f(y)}-\frac{b f'(y)}{f(y)^2}+O\left(b^2\right)$$ which make $$\frac{f(x+a)}{f(y+b)}=\left(\frac{f(x)}{f(y)}-\frac{b f(x) f'(y)}{f(y)^2}+O\left(b^2\right)\right)+a \left(\frac{f'(x)}{f(y)}-\frac{f'(x) f'(y) b}{f(y)^2}+O\left(b^2\right)\right)+O\left(a^2\right)$$ Ignoring the $O\left(a^2\right)$ and $O\left(b^2\right)$ terms, this reduces to $$\frac{f(x+a)}{f(y+b)}\approx \frac{\left(f(x)+a f'(x)\right) \left(f(y)-b f'(y)\right)}{f(y)^2}\tag 1$$ $$\frac{f(x+a)}{f(y+b)}\approx \frac{f(x) f(y)+a f(y) f'(x)-b f(x) f'(y)-a b f'(x) f'(y) }{f(y)^2}$$ Now, if we neglect the cross term $$\frac{f(x+a)}{f(y+b)}\approx \frac{f(x) f(y)+a f(y) f'(x)-b f(x) f'(y) }{f(y)^2}$$ which is exactly your last formula.
May I confess that I would prefer $(1)$ as an approximation ?