This part of the circuit is just a simple 555 astable with variable mark/space ratio.
The capacitor, along with the resistance values sets the frequency. The transistor is there to drive the mosfet which simply switches the current on and off through the primary of the transformer. The transformer steps up the voltage depending on the turns ratio.
The capacitor will be .01 microfarad (or 10nF as Andy correctly gives). By using a 100pF capacitor the oscillator will run at too high a frequency (100 times more than the original design).
I also agree with Andy that the transformer (and its design/construction) is your real problem.
My real concern about this question is that if your knowledge of electronics is this limited then building this type of circuit could be fatal. High voltage is not not an area a novice should tamper with.
The FB pin is an input that controls the operation of the TPS61165 switching regulator. If FB is less than 200mV the switching regulator tries to put more energy into the inductor to create a larger voltage on the output across C2.
If FB is greater than 200mV, the switching regulator backs-off transferring energy to the output capacitor C2 and the output voltage falls.
When there is 200mV on the FM pin, there is an equilibrium and the energy transferred to the output capacitor C2 is exactly balanced by the energy taken by the LED string.
What does all this mean? Well, to create 200mV at FB there must be \$\frac{200mV}{0.57\Omega}\$ amps flowing through the string of LEDs. That means there must be 351mA flowing through the LEDs.
This circuit is a constant curent drive to the LEDs, always ensuring that the current through them is about 350mA.
The internal circuitry is what is known as a boost-converter. Pin SW (means switch) and this is a mosfet that grounds the inductor for a short period of time. Energy into the inductor is accumulated in that short period of time and then the mosfet open-circuits and the inductor releases that energy, via D1 into the output capacitor C2.
FB stands for "feedback" and this controls the amount of energy accumulated in the inductor when the mosfet is grounded. This is usually done by shortening the length of time the mosfet is turned on. A shorter period of time equates to less energy in the inductor and less energy transferred to capacitor C2.
Best Answer
You're nearly there.
This is part of the problem.
While this is not wrong, it's not as useful as saying that transformers need voltage to induce voltage in their other coils. Currents do flow as well, but their relationship is more complicated than for voltage.
The feedback coil is in series with the base. This has the polarity (##) to drive the base to a higher voltage when the primary current is growing, when the primary has a positive voltage across it, and lower when it's negative. Once it starts to turn off, the feedback makes sure it turns off hard.
The resistor ratio makes sure there is enough voltage on the base to turn it on initially, when there's no voltage from the feedback winding.
The only other part of the puzzle is why the transistor should start to turn off anyway. There are two things that can do this, and the first to get there will trigger the end of the 'on' phase.
a) A transistor with base drive limited by R1 will only have a limited collector current it can support. Once the primary current has risen to this value, any further rise will pull the transistor out of saturation, and the collector voltage will rise, reducing the primary voltage. This will further reduce the base drive by transformer action to the base winding, and the transistor will quickly turn off.
b) The flyback core will magnetically saturate at a certain primary current. This causes the inductance to drop, which increases the rate of rise of primary current dramatically. It will now quickly exceed the transistor's limited collector current, whatever it was, and mechanism (a) will complete the switch-off.
(##) Thanks to Jonk for pointing out in comments that the polarity you might try to infer from that diagram is wrong. The absence of the explicit polarity indicating winding start dots should be a warning that this might be the case.