First draw the circuit with positive power at the top negative at bottom, power currents generally flowing down, and signals feeding left to right. If you do that, two useful things happen. First, many circuits will be drawn similarly most of the time, and you learn to recognize them after a while. Second, you will confuse yourself and anyone you ask to help you less in what is actually going on and what you hooked up where.
Redrawing the schematic so as to better illuminate the circuit, we have:
It is obvious why the LED doesn't come on, or blips on for a short time at best. That is because it is in series with capacitor C1. Capacitors block DC current. There can't be any sustained current thru the LED.
What you probably intended was something like this:
This allows the capacitor to be like a small reservoir for the LED. It will keep the LED lit for a short time after the transistor is shut off.
With this circuit you can see how a little base current can control a larger collector current, which is how a bipolar transistor is used to make circuits with gain.
However, the values don't seem right for what I think you want this circuit to do. Most LEDs are rated for 20 mA maximum, so R2 should be sized to that this can't be exceeded. Let's say it's a green LED and drops 2.1 V at full current, and that the transistor would drop another 200 mV. That leaves 9.0V - 2.1V - 200mV = 6.7V accross R2. From Ohm's law, 6.7V / 20mA = 335Ω, which is the minimum resistance to keep the LED current within spec. Therefore use the next higher common value of 360 Ω. That still results in nearly 19 mA LED current. You won't notice the brightness difference between 19 mA and 20 mA even in a side by side comparison.
Another problem is that there isn't enough base current to reliably light the LED to its full value. Let's say the B-E junction drops 600 mV, then there is 8.4 V accross R1, which results in 84 µA. Let's say you can count on a gain of 50, so the minimum LED current is only 4.2 mA. That's enough to see it light up on your desk, but not to reach full brightness. In reality, you will likely get a gain higher than 50, so you will get more LED current, but relying on that is bad design.
Let's work backwards to see what R1 should be to fully turn on the LED. Again we'll assume the transistor has a gain of 50, and we've already said the maximum LED current is about 20 mA. 20mA / 50 = 400µA. With 8.4 V accross R1 from above and using Ohm's law again, the maximum R1 value is 8.4V / 400µA = 21kΩ, so the common value of 20 kΩ would make this a nice and reliable circuit if the intent is to light the LED to full brightness.
What you have is a series RLC circuit. One of the most common introductory circuit in EE. For example, look under Series RLC Circuit in this wikipedia page:
en.wikipedia.org/wiki/RLC_circuit
These are equations copied from the same page for the under-damped case. Given R, C along with measured \$\omega_d\$, L can be solved from these equations. (You don't even need to measure the attenuation factor \$\alpha\$.)
If the RLC do want to oscillate, the node where D1 cathode is connected will go negative. Therefore, depending on how D1 is driven, it may conduct during the negative half cycle and kill the oscillation.
Best Answer
When the circuit is first turned on the voltage on pin 2/6 is low, this triggers the 555 which turns on pin 3. The 10n cap then charges through the 33k resistor. When the voltage on pin 2/6 reaches 2/3 Vcc the output turns off and the 10n capacitor starts to discharge through the 33k resistor into pin 3. When the voltage drops to below 1/3 Vcc, the process repeats.
When the 555 is triggered the NPN is turned on, the exact current allowed through the coil depends on the 1k resistor and which transistor was chosen. You can vary this by change the values of these two components or changing the timing values presented by the 10n cap or the 33k resistor.
Since the current through a coil cannot change immediately it slowly ramps up when the 555 is triggered. If the timing interval is too long the current can ramp up to an extremely high value. You thus have to make sure the current is limited either by component values or the timing.
Something else to take note of is that the voltage at the collector of the transistor can be extremely high (couple of hundred volts easy), it is pretty likely that you are blowing the NPN transistor after only a short operating time. I would recommend putting in some feedback at the output. Use a voltage divider on the output of the transformer, into an NPN transistor that pulls the reset line low (it should have a pullup). This will stop over-voltage killing your circuit.
So to summarize, there are three problems with this design:
As for the voltage decreasing over time, pulling 1A from a 9V will degrade its ability to supply power pretty quickly, this will drop the output voltage/current which affects the circuit. To test this you can monitor the voltage of the 9V battery while the circuit is operating, you should see it drop drastically.