Electronic – Does the rating of an AC fuse point to the amplitude or rms value of the current

The 180W reasoning is wrong. You don't have the full 12V across the fuse, only a fraction of it. Say the fuse has a 0.1\$\Omega\$ resistance, then 15A will cause a power dissipation of

\$ P = I^2 \cdot R = (15A)^2 \cdot 0.1 \Omega = 22.5W \$.

And that 22.5W in the thin wire is just too much, and the fuse will blow.
Note that the 12V doesn't appear anywhere in this, and you could do the same calculation for 15A at 230V AC. There's a "but", however. Fuses do have a rated voltage. Not for the blowing, as we've seen, but to make sure the remains of the fuse won't cause sparks when at a high voltage. So you can't use a 12V car fuse in a mains appliance, but the other way around should be safe.

Automatic fuses also react on current only. They have a coil which releases a latch if the number of coil-windings exceeds the rated value. Also here it's independent of voltage.

The current rating of a fuse represents the minimum sustained current the fuse will blow at ... eventually. A 1A fuse will take 1A for a very long time without blowing, and if the fuse can dump some heat into the PCB or has airflow across it, may never blow at 1A.

The critical parameter is the \$I^2 \cdot t\$ rating, which gives you an idea of the energy (power and time) needed to blow it. (Remember that fuses are really meant to protect circuits when there are catastrophic failures.)

It's crucially important to match \$I^2 \cdot t\$ ratings, since if you replace a fast acting fuse with a slow-blow type, even thought they both say 1A, it's going to take radically different energy levels to actually blow them.

When the fuse is intact, you only have an \$I \cdot R\$ voltage drop across it. This drop is going to be nowhere close to the voltage rating of the fuse (else it acts like a big resistor and limits the energy available to your circuit.) Once the fuse blows, the voltage rating comes into play, which represents how much voltage potential the open fuse can withstand without flashing over and re-energizing the compromised load circuit.