Relation between Amps and Thickness of wires
You are absolutely right. The amps flowing through any wire has to deal with cross-sectional area or lets say thickness of the wire. The relation between them is given by: I=VenA; I is current, e is charge of an electron, n is no. of free electrons, and A is the cross sectional area of the conductor. So increasing the area increases the current through the wire.

Another way to view this is through the idea of resistance. Since all the practical conductors have certain resistance, its resistance is expressed as
R=(r*l)/A
; r is the resistivity of the material, l is length of the conductor and A is the area. Here, you can see that the higher is the cross sectional area, lower is the resistance of the conductor. So more current can flow through the wire, thanks to Ohm's law.

Coming to your second question, the conductor with zero resistance doesn't heat. The heating of the practical conductors is due to their resistance. The heat energy is generated due to the loss of energy of electron in conductor. The electron can only lose energy if and only if it has some resistance in its path. So in case of ideal conductors, there is no any resistance offered, so there is no any way a electron can lose its energy in any form(heat energy, light energy, sound enery or whatever).