How is kinetic energy accounted for when describing diatomic bonds with a Morse potential?
The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation values of the total energy and potential energy.
Because the math is simpler I will limit the discussion to the harmonic oscillator (ie quadratic potential). For the QM harmonic oscillator the expectation value of the potential energy turns out to be
$$\left&lt; E_\text{pot} \right&gt; =\frac12 \left(v+\frac12\right)\hbar \omega=\frac12 E_\text{tot}$$
The expectation value of the kinetic energy is then
$$\left&lt; E_\text{KE}\right&gt;=E_\text{tot}-\left&lt; E_\text{pot} \right&gt;= \left&lt; E_\text{pot} \right&gt;$$
So it turns out that the kinetic energy and potential values have identical expectation values, consistent with the prediction of the virial theorem for a quadratic potential. This  is all outlined very nicely for instance in Atkins' physical chemistry textbook.
It is difficult to tell what energy levels are displayed on the diagram but they might very well correspond to the total energy given by $ E_\text{tot}  = \left(v+\frac12\right)\hbar \omega$.