Why a signal will take only GND plane on bottom layer for return signal?
It won't. If you have ground on both the top and the bottom layer, then you have this:
simulate this circuit – Schematic created using CircuitLab
Current will flow in both, proportionate to the resistances of each path. If the resistances are equal, the currents will be equal. If they are unequal, the lower resistance will carry more current. If they are very unequal, the much lower resistance will carry most of the current. Never does one path carry all of the current.
This is the same for AC signals, with the resistors replaced with inductors. Analysis of where the currents flow is complex, and will really depend on the specific geometry. However, with carefully controlled dimensions, this structure (ground plane on bottom, signal trace surrounded by ground plane on top) forms a transmission line called a grounded coplanar waveguide.
To your statements in the question, it's not exactly true that return currents flow exactly under the signal trace. This is only true for signals of infinite frequency. As the frequency gets lower than infinity, the inductance of diverging from this path becomes less significant, and the resistance of the ground plane becomes relatively more significant, so as the frequency gets lower, the return current will round the corners more. At DC, frequency is zero, so the inductance is irrelevant, and the current will prefer the lowest resistance path, which is probably the shortest path, if the copper is equally thick everywhere.