Hello, and welcome to Electro-Tech
You are correct in assuming that a resistor should be put in SERIES with the LED (between the battery and the LED). Assuming your 9 volt battery actually puts out 9 volts, and your colored LEDs draw 20mA, then for a resistor (for a single LED) you would need: (V=IR), 9=.02R, R=450 ohm resistor per LED. If you want to use two LEDs in series (which I recommend for the tail lights, etc., You could use a slightly lower value. If you use your 1K resistors, you would have 9=I(1000), I=.009, or 9mA, so the LED wouldn't be as bright, but should still be visible. Again, if you want two LEDs in series, you would almost definitely want a lower value resistor. It would be safe to assume you would need a resistors of about 500 ohms whether you want to put the LEDs in series or parallel.
Not quite sure what you mean by "Do I add them right into the circuit with however many are required placed into the positive connection and the equal amount into the negative connection?" One resistor for each LED (or pair of LEDs, as the case may be) is all that is needed. Simply connect your battery positive terminal to the switch, the other leg of the switch to the resistor, the other end of the resistor to the anode of your LED, and finally the cathode of your LED to ground (negative of the battery). This is a complete circuit and will allow you to switch your LED on and off with your toggle switch.
As for the math you may need, use Ohm's Law (which is what I used earlier--V=IR), where V is the voltage, I is the current in Amps, and R is resistance of your resistor. This is probably all you need for these circuits.
I'm afraid I'll have to let someone else draw the diagram for you, as I am rather busy at the moment. Let me know if you need me to clarify anything, though
Best regards,
Der Strom