Why is plane wave basis set is good for 1s and 2s orbitals, but for higher convergence is poor?
I guess by "higher" you mean higher principle quantum number $n$ (i.e., 3s, 4s-orbitals). For these orbitals, the radial wavefunctions have lots of wiggles near the atomic core (so that they are orthogonal to the other s-orbitals). To represent a wave function by a set of plane waves, you are essentially expanding that wave function using the Fourier series, and a sharp peak needs a Fourier basis (in this context, the Plane waves) with much higher frequencies to be accurately described.
As the plane wave (PW) basis can be systematically improved by setting a higher kinetic energy cut-off (which is related to the highest frequency component of the PW basis) I wouldn't say the convergence is a problem, you just need to include more PWs. The problem is the cost of storing and performing matrix operations with a huge amount of PWs.
As a side note, modern DFT codes utilize pseudo potentials with frozen core approximation so that only certain frontier valence orbitals are used in the actual calculations and their "wiggly tail" behavior near the core region is replaced with a smooth one (and the corresponding wave functions are called the pseudo wave functions) so that less PW is needed to describe them. This of course comes with a certain price but I will not dive into the details here.
Why is it better to use bigger number of k-points for case of unit cell relaxation (or any other calculations) and only one $\Gamma$ point for supercell?
This is related to the fact that the reciprocal space is the Fourier transform of the real space, so the cell vector in reciprocal space is inversely related to the cell vector in real space. For very small cells (e.g., unit cells), the lattice vector in reciprocal space is very large, thus we need a large number of k-points to sample the crystal cell in reciprocal space (Brillouin zone). However, when the crystal cell is very large in real space (e.g., large supercell), the Brillouin zone is very small, and usually $\Gamma$ point is enough to describe that cell.
Hope this helps.