Let us assume it takes 5 seconds to start the engine. During these 5 seconds the battery voltage drops to 9V. Let us assume the minimum voltage to operate the music player is 10V and at this voltage it draws 2A; thus its effective resistance R is 10/2 = 5 Ohms.
We can now imagine a capacitor charged to 12V before the engine starts and then being discharged over a 5 second period supplying the current for the music player during the cranking of the engine. During this period the voltage of the capacitor drops by 10/12 = 0.83 (I have used 10V to have a 1V margin over 9V).
In the attachment you will see the capacitor connected across a resistance R (the music player).
In such a simple CR series circuit the voltage across the capacitor Vc at time t is:
Vc = 12 x e (exp -t/CR) .... minus because the voltage is decaying.
Vc min/12 = 0.83.
Thus 0.83 = e (exp -t/CR) = e (exp -5/CR)
Therefore, ln(0.83) = -5/CR = -0.19
But R is 5Ohms.
5/5C = 0.19 (call it 0.2 for now).
1/C = 0.2 so C = 5F
Such a capacitor is a supercapacitor.
The affordable supercapacitors I found from RS only had a rated operating voltage of 3V dc.
The car's electrical system is 12V so I doubled this to suggest the required operating voltage of the supercapacitor ie:24V
When two identical capacitors are wired in series the effective end to end capacitance is halved.
Since the maximum voltage capacitor is 3V and 24/3 = 8 = 2 exp3, one would need to wire 8 supercapacitors in series. This means each SC would have to be 40F since 40 halved, halved again and then halved again (ie 1/2 exp3) is 5F.
MAL223091001E3 | Vishay 40F Supercapacitor EDLC -20 %, +50 % Tolerance, 230 EDLC-HV 3V dc, Through Hole | RS Components - https://uk.rs-online.com/web/p/electric-double-layer-capacitors/1802317/
The circuit then is as attached. The power diode is to isolate the circuit from the battery during the cranking of the engine. I have not yet done the sums on the required peak current handling of this diode and don't have time right now. Perhaps someone else could have a go?
I have not checked my sums but think I am right.
https://www.electronics-tutorials.ws/rc/rc_1.html