Using LinkSwitch AC-DC converter as shown.
Output voltage is 12V, 120mA.
Converting 12V to 5V using a regulator will dissipate a lot of power as heat.
Is there any way to adjust the component values to get 5V output?
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Right under the circuit diagram you used in your question it tells you: -
So with the values shown in your circuit, 1.65 volts is produced at FB with respect to pin S. For a 5 volt output you'll need to lower the 13 kohm resistor - you do the math. Also check this resource (Application Note AN-37 LinkSwitch-TN Family) to establish what inductance value changes are needed when operating at an output voltage of 5 volts.
Be also aware that this design is dangerous to the uninitiated - it doesn't provide any galvanic isolation from live AC voltages and can easily kill someone.
As the datasheet you linked says:
The values of R1 and R3 are selected such that, at the desired output voltage, the voltage at the FEEDBACK pin is 1.65 V.
R1, and R3 form a voltage divider, you need to work out values for them such that they generate 1.65 V at the feedback pin.