As Raskolnikov says, the first approximation is actually
$ \frac{2}{c}\left(1-\frac{2m}{r_1}\right)^{1/2}\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right) $
This is a valid approximation because the power series for $(1-x)^{1/2}$ is
$1 -\frac{1}{2}x+\cdots$
So as long as $x=\frac{2m}{r_1}$ is close to zero, the above approximation is a valid first-degree approximation. Expanding this substitution,
$ \begin{align} \Delta\tau&\approx\frac{2}{c}\left(1-\frac{m}{r_1}\right)\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right)\\ & = \frac{2}{c}\left(r_{1}-r_{2}-\frac{m(r_1-r_2)}{r_1}+2m\left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}\right) \end{align} $
So the second approximation that has been made is
$ \begin{align} \left(1-\frac{m}{r_1}\right)\ln\frac{r_{1}-2m}{r_{2}-2m}&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align} $
This is equivalent to the following approximation using logarithm rules
$ \begin{align} \left(1-\frac{m}{r_1}\right)\left(\ln(r_1)+\ln(1-2m/r_1)-\ln(r_2)-\ln(1-2m/r_2)\right)&\approx\ln\left(\frac{r_1}{r_2}\right)\\ \end{align} $
Now you just drop all the terms that have $\frac{m}{r_i}$, and your approximation is another logarithm rule. It is valid to drop these terms, because presumably $\ln(r_1)$, $\ln(r_2)$, and $1$ are relatively much larger.