According to physics, isn't recoil felt on both sides of the bullet?
Yes its true, but momentum (mass times velocity) is not very important by itself.
A .45ACP 230gr at 850 ft/s has ~28 ft-lb/s momentum. (7000gr to the lb)
A 9mm 115gr at 1200 ft/s has ~20 ft-lb/s momentum.
Each also has a recoil momentum component from the powder charge mass, but this is small enough to usually ignore for handguns, and not very different for 9mm vs. .45ACP.
What you "feel" as recoil is best described as "free recoil energy". The momentum of the bullet imparts a recoil velocity to the gun and the guns mass times that velocity squared is the "free recoil energy".
For a 2lb .45ACP pistol:
28 lb-ft/s divided by 2 lbs = 14 ft/s recoil velocity and (14*14*2)/(2*32) = ~6 ft-lbs free recoil energy. (the 32 is the gravitational constant to account for pounds force vs pounds mass, the math would be clearer in metric units)
For a 2lb 9mm pistol:
20 lb-ft/s divided by 2 = 10 ft/s recoil velocity and (10*10*2)/(2*32) = ~3 ft-lbs
Hence the claim the .45 has about twice the recoil of the 9mm
A 4 oz baseball at 90 mph is 132*0.25 = 33 ft-lb/s momentum and (132*132*.25)/(2*32) is about 68 ft-lbs energy -- we'd have little trouble catching it, which is also about the same muzzle energy of a .25ACP round which nobody would want to catch!
Beyond this, carry the most gun you can hit fast and accurately with would be my advice, but if fast follow-up shots were decisive, we'd all be carrying .22lr