I have been trying for a while to understand Coaxial Cable impedances for ham radio etc. use. I only know what it is not though.

Coaxial Cable does not measure the rated impedance from one end to the other

Coaxial Cable does not act like a load itself at the rated impedance, they may loose some power to heat, but they mainly carry the signal to the antenna (or whatever other load)
So, what is a coaxial cable impedance rating, and why can't I use a 75 ohm in place of a 50 ohm on the output of my transmitter?
Best Answer
Imagine, for a minute that electricity travels quite slowly.
When you turn your light switch on what happens? Current starts to flow  it starts working its way down the wire and so does the voltage. The current that flows is determined by two things: 
Will the current be too small or will the current be too much? It is the cable (and its properties) that dictate the amount of current flowing.
Voltage and current are traveling to the "unknown" load and because V and I are flowing there is power flowing (P = VI). When the current and voltage reach the bulb, if the bulb's resistance doesn't match the V/I relationship not all the power is consumed.
This means the excess (or deficit) of power has to be reflected back up the wire to the switch. It's got nowhere else to go.
In the real world of data comms or radio, this causes "reflections" and these can add or subtract to the forward power traveling down the wire and, in the case of data, it can become misshaped leading to possible data corruptions. In the case of an RF carrier, there will be points along the wire where it appears unmeasurable.
The cable dictates how much current initially flows based (mainly) on its inductance, capacitance and resistance. The formula is this: 
Characteristic impedance = \$\sqrt{\dfrac{R+j\omega L}{G+j\omega C}}\$
R is resistance per metre, L is inductance per metre, C is parallel capacitance per metre and G is parallel conductance per metre. At high frequencies (>1MHz) the impedance starts to largely become: 
\$\sqrt{\dfrac{L}{C}}\$ and if you look at some coax specs you'll see that 50 ohms is the result of this calculation.
Hopefully, by now you should be able to answer this.