I don't know why I wrote that sentence. It is not correct. Thanks for pointing out my mistake.

Regarding the square and the linear transformation, yes, what you describe is correct. But the term "uniform density" depends on the domain, not just the density function.
For the interval [0,1], the function f(x)=1 is the uniform density function. If I stretch the domain by a factor of 2, then the function g(y)=1/2 is the uniform density on [0,2]. You can read about how to convert one density into another under a linear transformation. For this example, with L=2 being the linear transformation, we have g(y) = (1/L) f(y/L).