Capacitor size across power source selector relay
You are more or less close to the answer, but you should rather use the capacitor energy equation: \$E = \frac{1}{2} C * V^2\$

The Energy that you need to store can be calculated as

\$E_{needed} = Power * time = ( 12 V * 10 A ) * 10 ms\$

Then you have two instants of time, before and after the switching:

\$E_{before} = \frac{1}{2} C * V_{before}^2\$

\$E_{after} = \frac{1}{2} C * V_{after}^2\$

then \$E_{needed} = E_{before} - E_{after}\$

Also you want \$V_{after}\$ and \$V_{before}\$ close to each other, perhaps 1 V is ok (you must check the load specifications), so there is enough voltage to feed the load, \$V_{before}=12 V\$, \$V_{after}=11 V\$.

The voltage will always go down a "little", that is the rule when discharging capacitors, how "little" depends on how big the capacitor.