In Exploring Black Holes, when dealing with the metric for polar coordinates for flat spacetime, he says "This derivation is valid only when dΘ is small - vanishingly small in the calculus sense - so that the differential segment of arc rdΘ is indistinguishable from a straight line." (page 3-2)
So just how small is vanishingly small? Lets say we are working at r = 1,000 metres from the event horizon. Can we reasonable work with a line 1 degree wide? or .1 degree? or .01 degree?
Your question pertains to the 2D Euclidean metric in polar coordinates. Given that $$x = r \cos \phi \qquad y = r \sin\phi$$ It's easy to show that the distance between any two points $(r_1,\phi_1)$ and $(r_2,\phi_2)$ is $$\Delta s^2 = \Delta x^2 + \Delta y^2 =\big(r_2\cos(\phi_2) - r_1\cos(\phi_1)\big)^2 + \big(r_2\sin(\phi_2) - r_1\sin(\phi_1)\big)^2$$ $$= r_1^2 + r_2^2 - 2 r_1 r_2\cos\big(\phi_2 - \phi_1\big)$$
If we define $\Delta r \equiv r_2-r_1$ and $\Delta \phi\equiv\phi_2-\phi_1$ and drop the subscript on $r_1$, then
$$\Delta s^2 = \Delta r^2 + 2r^2\big(1-\cos(\Delta \phi)\big) + 2 r\Delta r \big(1-\cos(\Delta \phi)\big)$$
This expression is exact for any value of $\Delta r$ and $\Delta \phi$. However, expanding about $\Delta \phi,\Delta r= 0$, this becomes $$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 + \left[ r\Delta r (\Delta \phi)^2 -r^2 \frac{(\Delta \phi)^4}{12} +\ldots\right]$$
The terms outside of the brackets are quadratic in $\Delta r$ and $\Delta \phi$, whereas the terms inside the brackets are cubic and quartic, respectively. If you want to know how good a particular approximation is, then you can evaluate the terms in the bracket and compare them to the terms outside.
Note that we can also rewrite the above expression as
$$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 \left[ 1 + \frac{\Delta r}{r} + \frac{(\Delta \phi)^2}{12} + \ldots \right]$$
The larger point is that if we let $\Delta r,\Delta \phi$ be infinitesimal quantities $\mathrm dr,\mathrm d\phi$, then lowest order terms the quadratic ones and we find that $$\mathrm ds^2 = \mathrm dr^2 + r^2 \mathrm d\phi^2$$ This can be used to compute the distances along any curve $C :[0,1]\ni \lambda \mapsto \big(r(\lambda),\phi(\lambda)\big)$ by integration:
$$\Delta s = \int_{0}^{1} \sqrt{r'(\lambda)^2 + r(\lambda)^2\phi'(\lambda)^2 } \mathrm d\lambda$$
