Your professor's premise of dissolution is only true if the process is incongruent melting which by the way is still melting. Whereby the process proceeds as:
$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][T1] KAl(SO4)2~_{(s)} + 12H2O_{(l)}\tag{$\beta$ to $\alpha$ + liquid}}$$
Then
$$\ce{KAl(SO4)2~_{(s)} ->[\Delta][T_2 > T_1] KAl(SO4)2~_{(aq.)}\tag{$\alpha$ to liquid}}$$
This concept is outlined in the generic binary phase diagram below where $\alpha$ is anhydrous potash alum, $\beta$ is the hydrated form, and liquid is potash alum solution. Above a certain temperature $\beta$ will decompose into $\alpha$ and liquid. As the temperature rises more potash alum will dissolve until eventually it will all be liquid again.

Figure 1. Binary Phase Diagram with Peritectic melting
However if it is a direct conversion from liquid to solid then it is congruent melting and your professor is wrong. This can occur either as melting an eutectic (shown in Figure 2) or congruent melting of a compound (shown in Figure 2) and proceeds directly as follows.
$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][92.5^\circ C] KAl(SO4)2~_{(aq.)} + 12H2O_{(l)}\tag{Interionic Melting}}$$
$$\ce{Sn_{(s)} + Pb_{(s)} ->[\Delta][183^\circ C] Sn_{(l)} + Pb_{(l)}\tag{Eutectic}}$$
$$\ce{Ti5Si3_{(s)} ->[\Delta][2130^\circ C] 5Ti_{(l)} + 3Si_{(l)}\tag{Intermetallic Melting}}$$

Figure 2. Binary Pahse Diagram with Eutectic.

Figure 3. $\ce{Si\! -Ti}$ Binary Bhase Diagram with Compound at $\ce{Ti5Si3}$ That Undergoes Congruent Melting.
Usually differential scanning calorimetry is used to determine which melting regime you are in indicated by a range of temperature causing a spike in heating rate corresponding to incongruent melting or a single temperature where heating spikes which would indicate congruent melting.
In this case, however, we can confidently assume that $\ce{KAl(SO4)2.12H2O}$ is under going incongruent melting since it melts above the temperature of water and below the temperature of the anhydrate and there are no other hydrates that it forms, the process is the peritectic melting thus the process your professor describes is correct, but it is still considered a melting process.