The only place you drifted off was in equating pressure to force. It is related, but is actually a compound unit of force per unit area. So pressure is in pounds per square inch, while force is just in pounds. 52,000 psi means that if you took any particular square inch of area inside the chamber and behind the bullet, it would have a total of 52,000 pounds force applied to by that pressure. Two square inches would have a total of 104,000 pounds force applied to it, etc.
So, your .260, including rifling, will have a bore cross-sectional area of about 0.054 in˛, while your 7 mm will have one closer to 0.063 in˛. At the moment of peak pressure that would be a force on the bullet base of 0.054 in˛ x 52,000 lb/in˛, while the 7 mm would apply a force of about 0.063 in˛ x 52,000 lb/in˛.
In both cases the square inches cancel each other out and you are left with 2,808 lbs pushing the .260's bullet base, and 3,726 lb pushing on the 7 mm bullet base.
Your resulting velocities don't reflect that full pressure difference ratio, but that's because of other factors. The bullet's total acceleration force is the average pressure it sees during its whole time spent in the barrel, minus friction losses. The pressure peak occurs at just one moment in the bullet travel down the tube. Differences in powder charge size and powder burning rates also come into play in determining the lower-than-peak pressures during the rest of the time the bullet is in the barrel, and not just where it was when the pressure peaked. Then there is the pressure gradient that develops because the gas has to chase the moving bullet, and that results in a few thousand psi less pressure at the bullet base than a pressure graph taken at the chamber shows. So, its complex, but you get some idea of what is going on.
Bookmarks