How to work to find the values of the components that should be used in a project
There is no general answer other than do the math and figure them out.  This of course requires truly understanding how the circuit works.  When you do understand it, knowing how to compute component values pretty much falls out naturally.

In your specific example, you start by understanding what each component does.  R2 sets the current to light the LED when Q1 is off.  I'm assuming the bottom of the LED is supposed to be connected to ground.  Let's say it's a typical T-1&frac34; green LED rated for 20 mA and 2.1 V when on.  Since 2.1 V will be across the LED and the power supply is 6 V, that leaves 3.9 V across R2.  From Ohm's law, (3.9 V)/(20 mA) = 195 &Omega;.  That's the absolute lowest value given the LED specification.  The common value of 220 &Omega; would work if you really want the maximum light.  Note that 200 &Omega; 5% is not OK since it could be as low as 190 &Omega;.  If you don't need every last bit of light, use 300 &Omega; or higher to save current, get longer lifetime from the LED, and be more tolerant of supply voltage variation.

R1 controls how much base current the transistor gets when the switch is on.  Figure the B-E junction will drop 700 mV, so that leaves 5.3 V across R1.  You didn't specify the transistor, but let's say it has a minimum guaranteed gain of 50 in this application.  The collector current needs to be up to 20 mA, so that means the base current must be at least (20 mA)/50 = 400 &micro;A.  Using Ohm's law again, (5.3 V)/(400 &micro;A) = 13.3 k&Omega;.  That's the maximum value that is guaranteed to work.  The common value of 10 k&Omega; therefore will be fine.

Added

I see from the comments it's necessary to dispell some myths about transistors used as switches.  There is no place in engineering for blindly following rules of thumb.  Conventional wisdom is often not the convention, and rarely wise.  There is no substitute for actually understanding what is going on.  When you do, you usually don't need the "every knows" rules.

Transistors saturate when the external circuit limits the collector current to less than the base current times the gain.  At this point, the C-E voltage goes to a few 100 mV and changes little with changing collector current.  You can flip this around and calculate the ratio of collector current to base current.  Saturation then means that this ratio is less than the actual gain of the transistor.

In the example above, I said the gain of the transistor was at least 50.  Therefore the transistor is in saturation for any collector to base current ratio less than 50.  In the example, I started with the collector current, then made sure the base current was more than 1/50 of that.  Same thing.

There is no magic C/B current ratio that must always be used for a transistor in saturation, or when it is used as a switch.  10 works for most transistors in most situations, but will also require more base current than necessary in many of those situations.  Sometimes you might use a fixed number like that when you don't know anything more about the transistor, but in this case we do know something more, which is that the actual gain is 50.  The base current being 1/30, for example, of the collector current is perfectly fine when you know the gain is 50.

There are some cases where you might want to deliberately overdrive a transitor way past what the gain tells you.  The C-E voltage doesn't change much with collector current or lower C/B current ratio, but it does still change.  Sometimes extra low saturation voltage is more important than more base drive current, and you overdrive the transistor to get a 100 mV or whatever lower drop:


In this case, a few 100 mV saturation voltage is no issue, so there is no reason to drive the transistor extra deep into saturation.  In this case, using 2 mA to drive the base wouldn't be a big deal either, so ending up with a C/B current ratio of 10 is not really wrong, but neither is ending up with 30 or 40.