I think this is exactly where people are getting confused. The input into the hydraulic system is still the same - therefore the sum of the outputs must be the same - unless you're affecting pedal travel distance as well.

I'll expand it into another concept bear with me on this, similar but maybe easier to visualize.
OK so a rope is pulled in tension between a U bar with tension of 1KN , that means that each end of the U bar at rope connection is resisting being drawn together with an equal but opposite 1KN force from each side, note 2 forces there not adding together.

Similarly if I attach a rope to a wall and I pull it away from the wall with 0.5KN (since covid I wouldn't manage a full 1KN) then I'm stretching the rope with a force of 0.5KN , and the wall is resisting in the opposite direction with a force of 0.5KN , and additionally between my feet and the wall the ground is being compressed by 0.5KN, now the crucial thing is the rope has no means to know (partly because its a rope) which end is the active force, as its being 'pulled' from both ends.

Ultimately you right though, being partly a semi intangible abstract concept there exists an explanation gap such that its becoming like we are trying to describe a colour in words.

The horse has already bolted, it applied a growing force to the stable door which initially resisted with an equal reaction up untill the latch failed at 4.3KN, adding brake pads and tyre compound at this stage would likely raze the stable to the ground.

That's a hair under 10N , but yes the force bearing down has an equal but opposite force bearing up from the table and all parts are being compressed with a force of just under 10N even though there are '2' 10Nish forces at play

Don't be daft. The force from the hydraulic pressure on two pistons is twice the force on one piston. Two have twice the area of one, and pressure is force divided by area. So if you have the same pressure and twice the area, you have twice the force. But this only applies to the two pot fixed caliper.

So in both case there is a compression force on the disc of 10N.

I'll take mild exception to the idea that there is something wrong with me, that's pushing civility a smidge what is ideally happening is that we are throwing ideas about till they stick in some manner.

Anything between two pistons is just a compressed strut by which force is transmitted from one to the other. 1N fed in one end is transfered into the other piston as 1N, however 1N from one side does not exist untill it has a 1N reaction.

You don't see the relevance of that despite the fact it is the same concept just the force roles were reversed for potential clarity nothing more.

But you'd need to push the pedal harder for that to be true. Are you suggesting that if I got 1000 hydraulic rams I could lift a house, but it'd feel no different than one ram lifting a feather? That's just not true. If that was the case then there would only be one size of hydraulic pump - tiny! You need larger input volume for more output volume - and in doing so you're requiring a larger force to act on it.

But that is only a complete description of a situation where there is only one force applied at one end of that strut. However, it is entirely possible to apply separate forces at both ends and have them combine. In which case, if there is an external force applied at end A and an external force also separately applied at end B, then the compression is the sum of the two forces. The resulting force at end B is the sum of the force applied there and the reaction to the force at end A. Just as the resulting force at end A is the sum of the force applied there and the reaction to the force applied at end B. At every point in the strut, and including at the two ends, there is then the sum of the force applied at A and force applied at B.

Graham