Why can copper have both four and six bonds?
$\ce{Copper(II)}$ complexes are $\ce{d^9}$ type complexes. This means that there are 9 d-electrons for the metal centre. When a complex forms, a number of ligands coordinate to the metal centre and the species adopts certain geometry. The five d-orbitals are no longer the same energy. They split into sets of different energies depending on the geometry and how ligands interact with them.

For an octahedral complex, orbitals split into two sets of degenerate orbitals - a lower $\ce{t_{2g}}$ ($\ce{d_{xy}},\ce{d_{yz}},\ce{d_{xz}}$) which is bonding to non-bonding and a higher $\ce{e_g}$ ($\ce{d_{z^2}}, \ce{d_{x^2-y^2}}$) which is slightly anti-bonding.

Copper is a bit funny as it experiences what is called a Jahn-Teller distortion where there is an elongation of the metal-ligand bonds along the z-axis. This is due to uneven occupancy of the $\ce{e_g}$ orbitals set. Then, the orbitals further split to remove the degeneracy and gain more stability. For the square planar complex of $\ce{tetraaquacopper(II)}$ to form you would need to go even further and remove those two ligands along the z-axis. The stabilization energy gained overall by going from octahedral to square planar configuration is what allows this switch to occur; it wouldn't happen if it were unfavorable. The gain in stability is borderline and that is why I assume that both configurations are possible.

Also, square planar fields tend to prefer stronger field ligands and water is not such ligand and this might also be considered as a reason why you get a mixture of the two types of geometry.