Emitter Follower Current Gain

The 0.7V drop is from the Base-Emitter junction being a PN junction (in an NPN transistor), which is the same as a diode (a silicon diode has a forward drop of ~0.7V).
A bipolar transistor is either NPN or PNP.

The reason it has current gain is that the base current turns the transistor on, allowing current from the collector (which is connected to V+) to flow to the emitter.

The reason it doesn't have voltage gain is due to the "negative feedback" effect from Re.
Let's run through an example of why the emitter stays 0.7V below Vb and does not reach V+.

Let's say we have this setup, and Q1 has a current gain of 200:

Now say we apply 3V to the base.
We know that the transistor begins to turn on when the base is ~0.7V higher than the emitter, so at this point current starts to flow from V+ into the collector and out through the emitter through Re to ground.
Now here's the important bit - when the current flows through Re a voltage appears across Re.

Now for arguments sake let's say the transistor "tries" to turn on fully and since we have a rising current flowing through Re, the voltage across Re rises also.

What happens when the voltage across Re reaches 2.3V?

Well, you should see where this is going now - the base is still at 3V. When the emitter was at 0V, the base-emitter (b-e) voltage was >0.7V and the transistor was on. Now, however, the b-e voltage is at 3V - 2.3V = 0.7V! so if the voltage across Re rises any further, the transistor would turn off. So the circuit has a natural limiting mechanism, and what happens is that it always sits at ~0.7V below the base voltage. It would not matter if the current gain is infinite, the emitter voltage cannot rise above this point without "turning itself off".

Here is a simulation of the above circuit, with the base voltage gradually ramped up from 0V to 3V:

Here's another simulation with a capacitor added in to prevent the emitter voltage from rising too quickly, so we can see how the transistor turns on fully at first to charge the cap as quickly as possible, then (almost) turns off again as the cap reaches ~2.3V and only the resistor current is left as things settle:

Simulation:

If you alter your diagram to look more like this it should work:

In your diagram, instead of a voltage divider and a capacitor you only have a single resistor. This won't work because the output of the Colpitts oscillator dips below 0.6 volts, and anything less than ~0.6V won't be picked up by an NPN transistor.

In the diagram, you can see that I've coupled the output of the oscillator with C2 to a voltage divider, with R1 and R2. The voltage divider adds a DC offset of 9V to the base of Q2. This way, the voltage at the base now ranges from 3V to 15 volts instead of -6V to 6V. Since the signal is always above 0.6V, Q2 passes it.

The output uses C1 to decouple the DC from the sine wave, leaving you with your original 6V amplitude sine wave, just with amplified current.

You can change the values of some of the components around if you want, but C2 should optimally be around 1/(2*pi*500KHz*R2), and C1 around 1/(2*pi*500KHz*R3).