and there was a typo as I am using 330 ohm resistor on both the red and green pins
No you have not ;) A 2 lead bi-color led has no "red lead" or "green lead". Both are connected to both leds inside. One is connected to the anode of the red and the cathode of the green. The other is connected to the cathode of the red and the anode of the green ;) And because only one of the two leds is on at the same time and it doesn't matter if the resistor is on the cathode or anode side you only need one resistor per led
....on each LED...standard practice....
First you talked about just paralleling each led. But a resistor per led is preferred. But how many do you connect? When you have 2 330ohm resistors per led the red will draw roughly 5mA. connecting 4 should not give a problem. But keep in mind that if you connect each led with a single 330Ohm (which is fine) you double the current.
The green might be dimmer because of the different voltage drop across it. Green leds come in two types. The classic (low-ish brightness) has a drop similar to the red. So when connecting it to the same resistor (as you do here, no matter if you use one or two resistors) it will draw about the same current. But modern high brightness green leds have a voltage drop like blue or white leds of around 3V. But that leaves 1V aka 33% less voltage for the resistor(s) then with a red led resulting in 33% less current for the green.
But I don't know which type of green you have (no led specs in this tread) so I'm not sure. It also just might be the green is just less bright.
A possible solution would be to increase the resistor for the red. But you can't simply attache a resistor only to red because of the bi-color 2-lead thing. But to do it, simply add a resistor in series with the one for green (doesn't matter on which side because it's a series connection). But you bypass that resistor with a diode in the current direction for the green. That way when you light the green the led sees only the non-by-passed resistor in series and for the red it sees both resistors (so the sum).
Only problem is, a diode has a voltage drop as well. So to make it work better you should use a Schottky diode.