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 base-emitter junction of a bipolar junction transistor is a diode:
simulate this circuit – Schematic created using CircuitLab
You have another diode in your circuit, the LED. And you obviously understand that you need a current-limiting resistor if you want to connect this diode to a 9V battery, because a diode has an approximately constant voltage drop that is less than your battery voltage.
The problem is the same with the transistor in your circuit. You have the base connected to the positive side of your battery, and the emitter connected to the negative side. You've done this, essentially:
simulate this circuit
What will happen here? A whole lot of current will flow until the battery can supply no more, or something melts. If you are using a 9V battery, it can't supply a whole lot of current, and I bet if you measure the battery voltage in your circuit, it's about 0.65V, unless the battery is dead by now.
So what you need is a current-limiting resistor on the base of the transistor, like this:
simulate this circuit
After you understand that, you might read Why would one drive LEDs with a common emitter?
Best Answer
Shouldn't it only light if I apply + voltage to the base?
No. If the base is disconnected (or the 1K resistor attached to the base is disconnected), is it still possible that the led lights up, even if there's no base current, due to leakage current in the transistor die, or due to the digital output from the eps8266-12 not being exactly at 0V.
It would be adviseable to connect a resistor (about 10k) from the base to 0V. This resistor provides a path for the leakage carriers to escape to ground, and this will stop the transistor conduction when there's no voltage applied to the 1K resistor.
Also, you're connecting the led incorrectly. The led connects the 12V directly to the transistor's collector, this is wrong. If the transistor conducts (if there is base current), the led will fry. You should put a load resistor (about 470 ohms) in series with the led. This is my suggestion of improvement to your circuit:
simulate this circuit – Schematic created using CircuitLab
Thanks to the user tlfong01 for pointing an error in the words that I used to describe the base-resistor.
Since English is not my home language, I ask for any suggestions to improve the text and thank you in advance.