Note you can't use the SUVAT equations here because they are only for constant linear acceleration, while in circular motion the direction is constantly changing. (In fact, really you should forget the SUVAT equations even exist: this is often one of the first things they tell you to do in first year undergrad physics. You should always do kinematics by integrating the equation of motion, though at A-level they write questions where it's much quicker to have just memorised SUVAT...).
To understand their argument, you just need to think geometrically. I think they have worded it as clearly as possible, so I don't really know what I can add, but I'll give it a go.
r vector rotates at rate
ω and it's magnitude is
r, so the magnitude of
v is
ωr. You should already know this.
Then, by the same token, since the
v vector rotates at a rate
ω as well, then the magnitude of the angular acceleration must be
ω times the magnitude of the velocity vector. Hence its magnitude is
(ωr)ω=ω2r, for exactly the same reason that the magnitude of
v is
ωr.
These geometric derivations are always tricky to get your head around. It becomes much easier to understand this when you do it "properly" (i.e. when you parametrise
r and differentiate each component and the basis vectors.)