Given a function:
$$B_t = 100e^{-r(T-t)}$$
We can perform a taylor series expansion for $r$ keeping $T, t$ fixed:
$$\displaystyle{B_t \approx (100e^{-r_{0} (T-t)})(1-(T-t)(r-r_0) + \frac{1}{2}(T-t)^2(r-r_0)^2)}$$
The author then says that the above can be divided by the first multiplier, i.e. by $(100e^{-r_0 (T-t)})$ to obtain:
$$\frac{dB_t}{B_t} = -(T-t)(r-r_0) + \frac{1}{2}(T-t)^2(r-r_0)$$
I cannot see how they get to this last line by simply dividing.
Essentially, the author is doing the following - $$B_t(r) = B_t(r_0) + B_t(r_0)(-(T-t)(r-r_0) + \frac{1}{2}(T-t)^2(r-r_0)^2)$$
Now, if $|r-r_0|<<1$, then $B_t(r) - B_t(r_0) \approx dB_t(r_0)$.
So, $$dB_t(r_0) =B_t(r_0)(-(T-t)(r-r_0) + \frac{1}{2}(T-t)^2(r-r_0)^2) $$
and the result follows.