How is the second capacitor charged in a Colpitts oscillator?
There are several ways to approach the Colpitts. Some work for some folks, others work for others, so I'll throw several at you and see whether any work.

In a parallel LC resonant circuit (that's not a DC circuit where you're charging capacitors steadily with a current, but in an AC resonant circuit where you need to take account of several cycles of oscillation), the tapped capacitor behaves as if it's an auto-transformer with a capacitor across it. A negative signal on the collector becomes a positive output on the base. Imagine if you will a grounded tap on the inductor instead, with the inductor behaving as an auto-transformer. But this transformation only works because the L and Cs are resonant. FWIW, the split inductor version of this circuit is called the Hartley oscillator, but with inductors more expensive than capacitors, it's usually the split capacitor one that is used.

Second way. Think of C1, L and C2 in that order as a discrete LC version of a transmission line, driven by the collector, with the base as a load. The electrical length of the transmission line is such that the signal inverts as it travels along it.

Third way. Ground is not really a 'signal stopping' terminal, it's merely a label carried by one particular node. The Colpitts Oscillator can be redrawn slightly to use the transistor in grounded base or grounded collector configurations as well. 

Here are some diagrams from the otherwise excellent wikipedia page on the Colpitts

The bias components have been removed to show only the resonant and gain components, and it's assumed that the positive rail is decoupled, so connected to whatever terminal is labelled ground. Why 'otherwise excellent'? Unfortunately, in those diagrams below, they've swapped the C1 and C2 component designators. It's the common base version that shares the same C1/2 labelling as your common emitter version. Spend some time redrawing them to see that they're all equivalent as far as signals go. 

One might spark your intuition better than another. It's the common collector one that works for me. I see the low impedance output of the emitter shaking the mid point of the capacitors, and getting voltage gain, in phase, at the 'top' of the resonator, driving the high impedance base.





Finally, simulate it in a SPICE simulation. Bear in mind that oscillator simulations usually have to be 'started' manually, usually by setting an 'initial condition' of voltage to the capacitors, to give it a kick. In real life, they self-start from noise, which is absent in the simulator. If you do that, start from a circuit with valid component values. Some ratios of C1 and C2 will not work. The loop gain depends on their ratio, and there has to be enough gain for the oscillations to build up. If you go to the wiki article and try to simulate figure 3, check out the talk page for discussion on whether it simulates or not. They discuss what initial conditions are needed, and how long it takes before oscillations build up.