Heating Nitinol while driving LED
You say you need 4.4W, at 0.35A inside the wire, so that makes 12.6V (rounded).

But, if your wire is only 2.9Ohm, that would mean (P = V^2 / R): 

V = sqrt(P * R) = sqrt( 4.4W * 2.9 Ohm ) =~ 3.57V

I = sqrt(P / R) = sqrt( 4.4W / 2.9 Ohm ) =~ 1.23A

I'll use my values for voltage and current, but leave you with the thoughts and formulae so you can redo it if you are 100% sure you were right about 0.35A (impying 12.6V for the wire, making about 24V for the total).

That means, if the LED module needs 12V, means the wire+LED should be powered with 12V + 3.57V =~ 15.6V, but 15V would probably do well enough.

You then need to adapt the LED end to handle the 1.23A, of course it normally wants only 20mA maximum, the module has an internal resistor that makes that happen for 12V, but if you power it through the wire with 15V, the wire will not add much resistance (2.9Ohm you say ?), which will cause the internal resistor to not limit enough any more, and the LED will in the best case be brighter with a slightly shorter life span (or die in the worst case, depends on how many chips and what resistor).

So you need to put an extra resistor in parallel to the LED that steals away 1.21A at 12V.

From R = V / I =~ 12V / 1.21A =~ 9.9 Ohm.

But that resistor needs to then be: P = I * V =~ 1.21A * 12V =~ 15W

Which means it will also dissipate a lot of heat, and not be very small at all.

If your wire really is 2.9 Ohm and needs 4.4W to get hot enough for your aims, you should seriously consider replacing the LED module with one that has no internal resistor, or is set for 3.5V or such. That would at least allow you to make the resistor:

R =~ 3.5V / 1.21A =~ 2.9 Ohm

P =~ 3.5V * 1.21A =~ 4.3W (still a lot!)

As you can see, in either case you are much better off adding a thin extra wire (possibly loosely wrapped around the Nitinol wire??) to power the LED separately.