I need help deriving the cutoff frequency of a negative feedback amplifier
I think they've missed out a couple of steps in between, so could
anyone explain how to get to those last couple of lines?

Just resolve the denominator: -
$$(1+\beta A_0)(1+ j\dfrac{f}{f_c(1+\beta A_0)}) = 1+\beta A_0 + j\dfrac{f}{f_c(1+\beta A_0)} +j\dfrac{\beta A_0f}{f_c(1+\beta A_0)}$$
$$= 1 + \beta A_0 + j\dfrac{f(1+\beta A_0)}{f_c(1+\beta A_0)}$$ $$= 1+\beta A_0 +j\dfrac{f}{f_c}$$
In other words it's correct.
For the 2nd equation, the only point of interest is this part of the denominator: -
$$1+ j\dfrac{f}{f_c(1+\beta A_0)}$$
This is the only point of interest because \$1+\beta A_0\$ (the bit of left out from the original equation) is just a constant gain modifier across the spectrum. And we know, that for a system like this, that to get the 3 dB point (aka \$f_{cc}\$), we equate real and imaginary terms thus: -
$$1 = \dfrac{f_{cc}}{f_c(1+\beta A_0)}$$
But, all of that could have been spotted from the very first equation you listed so I'm unsure why they drove round the hills to get there.