The first thing you have to note is that: no, the transverse $M_{xy}$ component does not exchange energy with the longitudinal $M_z$ component for the magnetization $\vec{M}$ to return to equilibrium. Remember that the energy of a magnetic dipole in a static field $\vec{B}=B_z\hat{z}$ is given by
$$
E=-\vec{\mu}\cdot \vec{B}=-\mu_zB_z=-{M_z}B_zV
$$
where $V$ is the volume of the sample (assuming the dipole distribution is uniform over the sample, for simplicity).
This should tell you that the energy of the sample only depends on $M_z$. The laws of physics do not require that the sample "regains" energy it "converted" to $M_{xy}$ as it relaxes, as you initially assumed. To address your question, the $T_2$-decay does NOT involve any energy transfer as the longitudinal alignment with respect to the static field does not change.
Indeed, the longitudinal $T_1$ and transversal $T_2$ relaxation occur independently through two different mechanisms. These mechanisms can even happen in totally different time scales. For example, the $T_2$ relaxation of a nitrogen-vacancy center happens within microseconds, while the $T_1$ relaxation takes milliseconds to finish. They are NOT dependent on each other, but they happen simultaneously. Once you tip $\vec{M}$ from its initial equilibrium (i.e. into the $xy$-plane), the following processes work to return $\vec{M}$ to the equilibrium.
The sample never loses any energy when you tip it toward the $xy$-plane. Instead, it gains energy from the control pulse you use on it. As the sample is no longer in equilibrium with its surroundings, it can "share" its extra energy (which it gains from the control pulse) with its surroundings via the $T_1$ relaxation. The "surroundings" here are usually the surrounding lattice, hence the name "spin-lattice relaxation". This is where the energy transfer occurs.
Meanwhile, the $T_2$ relaxation does not involve energy transfers. Due to inhomogeneity in magnetic fields, different parts of your sample may experience slightly different static field strengths. Now, you can break $\vec{M}$ down into the "constituting" magnetization of each part of the sample. By "part", I mean a chunk of your sample that experiences the same magnetic field strength but is different from the other parts of the sample. Each constituting magnetization undergoes Larmor precession at slightly different rates. Give it long enough, and they are dephased far enough from each other that they cancel out each other. This is what goes on during a $T_2$-relaxation---call it a "transversal component self-annihilation", if you will.
In conclusion, the decay of $M_{xy}$ does NOT result in the recovery of $M_z$. $M_{xy}$ decays because its constituting components are dephased from each other and cancel each other out, while $M_z$ recovers by giving its energy to its surroundings. The two processes happen simultaneously, independent of each other, and one process can finish faster than the others (as shown in your figure). The $T_2$ relaxation never involves energy emissions.