It's not a proper voltage regulator, but a potential divider (R50 & R49) with an emitter follower or buffer transistor (Q13).
VOUT = R49*VIN/(R50+R47) - VBE
VBE is the base-emitter voltage of the transistor, which will vary from 0.6V unloaded to around 0.8V, when it's passing a decent load current.
The above formula gives around 9.4V fully loaded. Changing R50 to 1k8 should give a similar output voltage with 12V in.
If you want to use the LM7809, then Q13, R50, R49, R4 and C4 will need to be removed. You might want to consider the LM78L09, which has a lower current limit of 100mA, which might be more suitable, than the 1A offered by the LM7809. Another option is the LM317L, which does require a couple of resistors, but gives better performance than the LM78L09.