Voltage regulator UA78S09 (9V/2A) supplying only 0,4A of current
If the voltage is 9V the regulator is doing its job. Its job is to output 9V (give or take) when the load is somewhere between 0 and maximum current (provided you don't let it overheat). It is doing that.  

The motor will draw more current if you put a load on the shaft, but it's not the regulator's job to force current through the motor. 

From your comments, I conclude that the 9V is not what the regulator is outputting when the motor is not moving. A quick look at the datasheet indicates that this particular regulator has foldback current limiting, so the output current will drop to approximately 500mA typically on overload (with 27V in, so 400mA is quite plausible with only 12V in). 

So it's likely unsuitable for driving your particular motor where the start-up surge exceeds the built-in current limit. 

There are several circuits shown in the datasheet which I linked above showing how to increase the current capability. The simpler one does not have short-circuit protection so testing it into a short circuit might have ill effects. 



You will be running pretty close to the dropout voltage with such a circuit however, so it might not maintain 9V out. 

Don't leave the recommended capacitors out! A bit higher values won't hurt, but they are recommended for good reasons, even though it's not your problem in this case.