That looks right - negative side of all power supplies tied to eachother and to Arduino ground. Note that as buck converters are not isolated, you don't need to explicitly connect each one's negative side to ground as long as the negative side of the battery is tied to Arduino ground. The negative side of of input and output of buck converter is internally connected.
You need at least 7 volts into the barrel jack due to the drop of one diode and the dropout voltage of the 5 volt regulator. 8 or 9 volts would be fine and insure some supply headroom since you have no load on the uno's 5 volt rail.
Also, you might want to consider changing the output of the buck converter for the arduino to the Vinpin and increasing the voltage to 9V so the buck converter does not have drop the voltage all the way from 12V to 5V because the arduino onboard regulator can drop it from 9V to 5V.
You should have about 100-150 ohm in series with the mosfet gate to limit the output pin current to under 40 ma at turn on, the mosfet needs to be a logic level device for full current output to the laser.
FYI, here's an example of how to draw your schematic so that it is easier to understand and follows schematic convention:
You can drop the 9V buck converter, and instead power the Arduino from the 5V buck converter. Saves one converter, saves risking overheating the regulator. 5V to the 5V pin and you're done. That Vin/barrel jack is only for convenience when prototyping, the moment you go to a semi-permanent installation you don't use it any more. And you should consider a Nano instead of an Uno for easier soldering.Do make sure the 5V buck converter can produce the rated current of the servos several times over (stall current!), and add a big decoupling capacitor.If that #3 is also a 5V converter, better use that for the Arduino, and leave the motors on the separate one.