Why doesn't the bidentate ligand form dative bonding three times?
Orthocresol already pointed you towards the Jahn-Teller effect. The effect is there in $\ce{[Cu(H2O)6]^2+}$, the hexaaquacopper(II) cation, but more pronounced in the tetraammindiaquacopper(II) cation $\ce{[Cu(NH3)4(H2O)2]^2+}$. Compare an image of the complex as shown below:



Even the naked eye can see that the copper-oxygen bonds are much longer than the copper-nitrogen ones. Measuring the distances in crystal structures containing that cation gives $203~\mathrm{pm}$ for the $\ce{Cu-N}$ bonds and $251~\mathrm{pm}$ for $\ce{Cu-O}$.[1] Why is this the case? To answer that we need to refer to the molecular orbital scheme of a typical octahedral complex:[2]



The three lowest groups of orbitals containing six MOs in total are typically populated by the ligand lone pairs which form the coordinate bond. The next set of two groups — five orbitals in total — are the metal d-orbitals that can (or cannot, in case of $\mathrm{t_{2g}}$) interact with the ligands. For copper(II), we need to fill in nine metal d-electrons meaning that the $\mathrm{t_{2g}}$ level is fully populated while $\mathrm{e_g}$ has three electrons in two orbitals — a denatured state.

This denatured state is not energy-optimal. Energy could be released if we somehow manage to unequalise the $\mathrm{e_g}$ orbitals, give them different energies and have the lower-energetic one be fully populated. Remember that the $\mathrm{e_g}$ orbitals are $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. In a gedankenexperiment, remove the two ligands in $z$-direction symmetrically away from the metal centre. Any orbital that has a $z$ contribution should now be stabilised — most notably, that includes $\mathrm{d}_{z^2}$. The orbital scheme looks slightly different, the complex’ symmetry is no longer $O_\mathrm{h}$ but $D_\mathrm{4h}$ but we have gained energy by being able to differentiate between the $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ orbitals. This is the Jahn-Teller effect.

Consequences of the effect include that ammin ligands preferentially coordinate copper(II) equatorially, and that the tetraammindiaqua complex is the preferred one in high aquaeous ammonia concentrations. Hexaammincopper(II) can be prepared only in liquid ammonia to the best of my knowledge.

Now let’s consider using $\ce{en}$ (ethylenediamine) rather than $\ce{NH3}$. The bridge between the two amino groups is long enough to bridge across one edge of the octahedron but not any further. Replacing all the equatorial ligands with $\ce{en}$ makes sense because we will get another very symmetric complex $\ce{[Cu(en)2(H2O)2]^2+}$, and because these are the ones that are replaced first. If we now wanted to replace the two other water molecules, we would create a rather distorted $\ce{[Cu(en)3]^2+}$ complex, where two $\ce{en}$ molecules have one short, strong and one long, weak bond. This is an unfavourable condition, since the former shorter, stronger bonds have to be weakened to get there. In short, $\ce{en}$ behaves like ammonia and only occupies the equatorial positions, not the axial ones. A consequence again of Jahn-Teller.



[1]: Data taken from Professor Klüfers’ internet scriptum for his general and inorganic chemistry course at the LMU Munich (section 23.4).
[2]: First presented in this answer and originally taken from Professor Klüfers’ coordination chemistry course.