Picture this in your head: the electron wavefunction is sinusoidal function spread over the entire molecule. In the LCAO-MO approach, the molecular orbital wavefunction is approximated as an array of atomic orbitals - each originating on one of the nuclear positions - that blend together. The orbital coefficient reflects the magnitude of contribution of each atomic orbital and can be visualized as the intensity of the wavefunction at the point in space where the nucleus is. So if you can picture in your head what the wavefunction is and how it is spread over the nuclei, you could, in principle draw or specify the relative magnitude of the orbital coefficient.

This is easiest for linear molecules in which the molecular orbital wavefunction is a single dimension function. The upper figure you posted basically tells you how to do this. If you draw the appropriate sin function (the wavefunctions of a 1D particle in a box are basically just sin functions – and make sure they are normalized) on a piece of paper, with the nuclei evenly spread nuclei underneath it, you can use that as a guide to draw how big the atomic orbital should be. The wavefunction extends beyond the terminal positions, keep that in mind. No math required, assuming you know how to draw a sin function well. (You can even quantify it using the equation provided there – note that you cannot just use vanilla sin functions to do this; there is a coefficient out front that arises due to the requirement that wavefunctions – and orbital coefficients – be normalized.)

This becomes quite a bit more difficult with nonlinear systems, in which the molecular wavefunctions are two dimensional and more complicated in form – although in principle the same idea holds.

However, this does all kind of miss the point that there is no one right set of orbital coefficients. The orbital coefficients are tied to the energy eigenvalues and the approximations one makes to the integrals mentioned in the earlier post. If you change those, the orbital coefficients can change quite a bit in magnitude.

(Apologies for the deleting an earlier post. I wasn’t satisfied with my original response.)