One resistor in series and two in parallel for LEDs
No, this is fine. Because you have a resistor in series with each LED this takes care of any kind of problems that might arise from matching. All LED's are not created equal, the I-V curves have a tolerance. It is probably 'slightly' better to have the LED's go directly off of the rail but this configuration works. The most important thing is to not have LED's directly in parallel.


  Also (just to double check I am doing my calculations right as I am
  still a student learning this stuff), assuming each LED has a forward
  voltage of 2V, is each LED drawing about 20mA here (7V across
  resistors, Rtotal = 355Ω so 7V / 355Ω = 0.0197 Amps)?


The drop across the shared resistor would be 220*40mA= 8.8V 
Something is wrong with your calc because this would mean that you would have more than a 9V drop total.

I'd just solve for the node voltage 

for the 220Ω resistor you get

\$\frac{9V-x}{220Ω}=i_{220}\$

Each LED would be 

\$\frac{x-2V}{270}=i_{LED}\$

The whole equation would be

\$\frac{9V-x}{220Ω}=\frac{x-2V}{270}+\frac{x-2V}{270}\$

Solving for x you get 4.66. Then solving 

\$\frac{9V-4.66V}{220Ω}=0.0184A\$

\$\frac{x-2V}{270}=0.0985A\$

or half of the amount you calculated through each LED.