BUCK converter considerations in a battery charger,with battery as a load
You terminology is reversed. A synchronous buck with the second active switch has the potential of less loss because the switch can have lower voltage drop than a diode.

Copied from your comment -- L = (Vin - Vout)/(Vin * frequency * r * Iout_max). r would be the inductor ripple current ratio. So (r * Iout_max) would be the inductor ripple current. I would say set r to the lower ratio unless the inductor gets too large. Notice that when Iout_actual (if you need to throttle the output current for some reason) drops below (0.5 * r * Iout_max), for a synchronous buck, the inductor current actually reverses; for a non-sync buck, the inductor current becomes discontinuous.

You do want an output capacitor for at least these reasons -- to prevent reverse current to the battery in case of synchronous buck and throttling the current is part of the scheme; to reduce the inductive effect of the output wires and load; to reduce EMI -- because otherwise the still relatively large inductor ripple current would go straight out. The output capacitor helps to reduce the ripple current going out to a fraction of the inductor ripple current.
If the battery is always floating then do which ever is easier, which is usually dictated by the regulator IC that you use. If the battery is connected, then you may have to use high side shunt to avoid a ground loop issue.




New edit to add info on output capacitor and output ripple current:

$$ I_{out\_ripple} = \frac{Z_c}{Z_{load} + Z_c} I_{ind\_ripple}$$

So the first order estimate is just a simple current divider relationship, where \$Z_{load}\$ would be the ESR of the battery in this case, and \$Z_c\$ the impedance of the output capacitor:

$$ Z_c = \frac{1}{2 \pi \times frequency \times C_{out}} $$