We have an equation of a Circle $$x^2 + y^2 = 1$$ and equation of a Cube $${||x-y|+|x+y|-2z|+||x-y|+|x+y|+2z|=1}$$
When I substitute x,y in first equation or x,y,z in the second equation then the equations are said to be satisfied when LHS and RHS are equal.
I have a quick question. If the first equation or second equation get satisfied, because LHS becomes equal to RHS, then those values of x and y of first equation are said to be lying on the circumference of the circle and not inside or outside of the circle and x,y,z of the second equation are said to be lying on the surface of the cube and not on the inside and not on the circle of the cube. Correct?

Yes. In set-builder notation, one would say the following
The circle $S$ consists of of $$\{(x,y) \in \mathbb R^2 \mid x^2+y^2=1\}.$$
And same goes for the second one.