The solar panel will only give out 560 mA max. If you are using a buck converter you are expecting to take the full output current, which could be nearly 1.6 A, some of the time and zero current the rest of the time. You need a capacitor, most likely electrolytic or tantalum, that can supply the current for the time needed, and then charge up when the buck converter is in the off period.
If we assume 100% efficiency, the output current will be 34*0.56/12 = 1.586 A. If you are running at 20 kHz, the period is 50 μs. The time that the output is on will be 12/34 = 35% of the time. The capacitor has to support 1.586 - 0.56 A for 35% of 50 μs, so the charge is just over 6 μC. With the 22 μF capacitor suggested, the change in voltage will be just over 0.25 V, so that looks fine, but the ESR (equivalent series resistance) of the capacitor will also have to be considered. There is no disadvantage in having a much larger capacitor.
If we assume 100% efficiency, the output current will be 34*0.56/12 = 1.586 A. If you are running at 20 kHz, the period is 50 μs. The time that the output is on will be 12/34 = 35% of the time. The capacitor has to support 1.586 - 0.56 A for 35% of 50 μs, so the charge is just over 6 μC. With the 22 μF capacitor suggested, the change in voltage will be just over 0.25 V, so that looks fine, but the ESR (equivalent series resistance) of the capacitor will also have to be considered. There is no disadvantage in having a much larger capacitor.