If you look at the derivation for the angular frequency it is as follows:
When looking at my notes it said that this frequency is also equal to that of the voltage (since the p.d must change directioons in order to keep accelerating the particle in its direction of motion)
Surely this angular frequency must be double this considering the voltage must change each time the particle makes a half turn, (rather than a complete revolution)
Any help would be appreciated
No. There are two voltage changes per period of the driving voltage, just the same as there are two half-turns per revolution of the particle. The period of the driving voltage matches that of the particle, so the frequencies must also match.
So we know that the cyclotron orbital period $T = 2\pi m / q B$ is independent of radius, and we know that each half takes time $T/2$ to clear. However we also know that an electron accelerates by going from a low voltage to a high voltage, and so when it's going from half #1 to half #2, we want half #2 to be at the higher voltage; when it's going from half #2 to half #1, then we want half #1 to be at the higher voltage.
This means that if you just arbitrarily choose half #1 as "ground," you will see half #2 as a square wave going from $-\phi$ to $+\phi$, and returning to where it was after a period $T$. If you used $T/2$ as this square wave's period, as you're suggesting, then after spending $T/2$ in the one half, the electron has now flipped its direction, but the square wave must return back where it started (because that's what "period" means for cyclic things). This means that the electron must be now fighting against whatever voltage helped it out earlier, since the voltage gradient is the same but the electron's direction has changed. So you really need this period to be $T$ so that you get the "up" ($+\phi$) half of the square wave one way and the "down" ($-\phi$) half the other way.
Converting $T$ into an angular frequency $\omega = 2\pi~f = 2\pi/T$ gives the result they quoted.
