An analogy may help to visualize this:
Think of the transistor as a valve or faucet. The base is the knob, the water tends to flow from the positive side (storage tank) to the ground (drain), if you follow the normal "current flow" directions.
The LED is like a little transparent glass section in the pipe, with a small ball loosely held in that section.
When the faucet is opened, water will be allowed to flow, and the little ball will jump around due to the water's flow.
This will happen whether the LED is above or below the faucet section.
Now for the case of electron flow, as opposed to conventional current flow direction.
Consider the same pipe and faucet, but with the ground being a source for some gas, say natural gas at high pressure underground.
The Vcc is the open air, normal barometric pressure.
Again, as the faucet is opened up, the gas will flow up the pipe, the little ball will bobble around. Again, the glass pipe section (LED) could be before or after the faucet, it won't matter.
I hope this analogy helped.
The transistor is in reverse active mode. The collector acts as the emitter and the emitter as the collector. This is possible because NPN reversed is still NPN.
The performance is worse than in forward active mode, because emitter and collector usually have different doping levels and a different structure. Therefore this mode of operation is not very useful.
Another problem with this circuit is that one has to be careful not to exceed the maximum reverse voltage of the base-emitter junction.
Best Answer
Your question says "current source" but I think you're really asking about switching voltage to a load.
With a emitter-follower you're stuck with around 600mV of voltage drop (plus whatever the driving circuit drops when supplying the base current) unless you have a higher voltage source available. So if you are driving an LED and have a 3.3V power supply and a GPIO that has 2.9V on it when driving base current, you're going to get about 2.3V to your LED, which will be rather limiting, plus the transistor dissipates power because of the current and voltage lost.
If you use a saturated transistor (PNP or NPN) then your voltage drop across the transistor might go down to a few tens of millivolts, and almost the full supply voltage is available for the load, and the transistor stays cool.
On the other hand, there are circumstances favorable to the use of an emitter follower, for example if you wish to draw current from a higher voltage (perhaps unregulated) supply and provide a relatively constant voltage (when 'on') to a load from that.
Most of the simple questions here don't require that kind of circuit, but I've certainly used it to benefit in commercial products where saving a bit of money can make a big difference.