Are phase transitions reversible when performed at supercooled temperatures?
The entropy change for a system is the same whether or not the process that takes the system from its initial to final states is performed reversibly. That's because the entropy is a state function. The difference between a reversible and irreversible process, all else being equal with the terminal states of the system, lies in what happens to the surroundings. 

Therefore, to find the change in entropy of the system you start by describing a reversible path for which you can compute the entropy change. Such a path might consist of the following steps: 

(1) Warm supercooled water from the initial temperature ($\pu{-20^\circ C}$) to the normal freezing point ($\pu{0^\circ C}$)

(2) Freeze the water to ice at the normal freezing point ($\pu{0^\circ C}$)

(3) Cool the ice from the normal freezing point to the final temperature ($\pu{-20^\circ C}$)

The path involves reversible heating and cooling of supercooled water  and ice, respectively, and the intermediate step of freezing water to ice at the normal freezing point of water ($\pu{0^\circ C}$). 

You can use the heat capacities to compute the entropy changes during heating and cooling of the system, respectively, from the initial and to the final states. For one mole of water:

$$\Delta S_{m,T_i\rightarrow T_f}=C_{p,m}\ln\left(\frac{T_f}{T_i}\right)$$

making sure to use the appropriate heat capacity (assumed constant over the temperature range) for water or ice.
For the first step, you warm up the supercooled water, and compute the entropy change by assuming that the supercooled state is stable. For the final cooling step, you compute the entropy change of the cold ice.

For the intermediate freezing step 

$$\Delta_fS_m=\frac{\Delta_f H_m}{T_f}$$

The molar change in entropy for the system undergoing the irreversible freezing process is then just equal to the sum of the above changes for the three steps. 

As for the computation of the entropy change of the surroundings: you compute the enthalpy change for the alternate path; then, since the enthalpy change for the actual irreversible process is the same as that for the reversible path, and the process is isothermal and isobaric (the intial and end states have same T and p), you can equate the enthalpy change with the heat, and can compute the change in entropy of the surroundings by dividing the heat by the temperature, making sure to use the correct value ($\pu{-20^\circ C}$).