No - sorry - you are in error...
You are allowing the math to interfere with your understanding of the actual physics.
In BE physics...
The equation given represents the total "pull" that is
between two objects - where the mass of both objects affect the outcome. However, the actual "pull" of each object
on the other is dependent
only upon its own mass.
The "pull" of a drop of water falling from a cloud above the ocean against the earth, the ocean, a ship on the water, etc is not the same as the "pull" of [each of] those things against the drop of water.
The greater mass has the greater "pull" on the other object.
The "pull" of the earth on that drop of water is greater than the "pull" of the moon on that drop of water.
And,
the overall "net" force on the drop will govern - it falls to earth - it does not "fall" to the moon.
The same is true for every drop of water that is in the ocean. The "point-blank" distance plus the mass of the earth does not allow for the moon - with much lower mass and at such a distance - to "overtake" the earth and "lift" so much mass/weight several feet.
If the earth has such greater "pull" on the drop of water than does the moon - how can the moon possibly "lift" so many more "drops" -
all in molecular cohesion from the 'local' forces that are in-effect - against the point-blank effects of the earth and the ocean? (Not to mention the other forces present - the ocean surface is very rarely glass-smooth, no water currents, etc.)
The "net" force is always too great in the earth's favor - the moon "doesn't have a chance" to produce that great of an effect on such a large mass of ocean.
And - the "excuses" for the 'bulge' on the side opposite the moon --- that is purely preposterous...
If the moon were causing the tides, then 'high tide' would only exist in one 'region' on the face of the earth at any moment in time - the mid-point of which would be the point closest to the moon at that point in time.