nichrome wire not heating up
$$V=IR$$

If your Nichrome wire is \$31\Omega\$, and you apply \$12\mathrm{V}\$, then you will have a current of \$387\mathrm{mA}\$ flowing through the wire.

If instead you place a \$330\Omega\$ resistor in series, you now have a resistance of \$330+31=361\Omega\$. So doing the calculation again, you now have only \$33.2\mathrm{mA}\$ flowing through the circuit. The vast majority of the voltage is now dropped over the resistor.

$$P=IV=I^2R$$

You have \$33.2\mathrm{mA}\$ flowing through your circuit, so the Nichrome wire is dissipating \$P=I^2R=33.2^2 \times 31= 34.2\mathrm{mW}\$.

On the other hand, the resistor is dissipating \$P=I^2R=33.2^2 \times 330= 364\mathrm{mW}\$. Basically 10 times as much.



So the question really is, why do you have the resistor?



If you instead place the resistor in parallel with the Nichrome wire, it would do nothing useful. The Nichrome branch of the circuit will still show \$387\mathrm{mA}\$ flowing. You will also have \$36\mathrm{mA}\$ flowing through the parallel branch of the resistor, but this wouldn't change the amount of heat generated in the Nichrome - for all intensive purposes the resistor is essentially a separate circuit sharing the same supply.



It seems that the issue is your supply cannot deliver enough power. The voltage of the supply is basically dropping away meaning that you are only getting \$\approx 160\mathrm{mA}\$ through the wire, less than half of what you need. Noting the square in the Power formula, this means that the power dissipated by the Nichrome is less than 20% of what it would be if the voltage had not dropped away. Try to find a supply which can deliver more power.