Why do ligands approach along the axes? According to the Crystal Field Theory (CFT), it is given that ligands (their lone pairs) approach the central metal along the axes. When they approach, the degeneracy of the $\mathrm{d}$ orbitals is lost. The explanation:

The $\mathrm{d}_{x^2-y^2}$ and the $\mathrm{d}_{z^2}$ orbitals(the $\mathrm{e_g}$ orbitals) are oriented along the axes while others are oriented in between the axes. In octahedral geometry having complexes, when the ligands approach, they approach along the axes and thus the $\mathrm{e_g}$ orbitals which are along the axes experience more repulsion. Hence, these orbitals increase in energy more than the $\mathrm{t_{2g}}$ orbitals. Thus, the orbitals split in energy due to the approach of ligands.

My question is  why do the  ligands (their lone pairs) have to approach along the axes. They can approach from anywhere.
You basically said it.


  They can approach from anywhere.


Well, I can define the axes to be anywhere.

An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. Hence, the $\mathrm{d}_{z^2}$ orbital does not have the ‘orientation’ you know from pictures — it only gets that once in some way axes are defined. You could now go ahead and define that one of your octahedral ligand approaches the central metal (placed at $(0,0,0)$) from the coordinates $(1,1,1)$. Get out a large sheet of paper and a new pencil and do the maths, calculating what the final orbital distribution will be. The maths will be a lot more complicated, but the result will be the same. Hence, chemists decide to not do complicated maths and just have octahedral ligands approach along the coordinate axes.

Note that in tetrahedral complexes, the ligands would indeed approach from $(1,1,1)$ and the symmetry-equivalent positions of a tetrahedron.