Why does the isomorphism of varieties “lines intersect / don’t intersect” argument work?

https://en.wikipedia.org/wiki/Rational_mapping

The usual example is that $ \mathbb {P} _{k}^{2} $ is birational to the variety $ X $ contained in $ \mathbb {P} _{k}^{3} $ consisting of the set of projective points $ [w:x:y:z] $ such that $ xy-wz=0 $, but not isomorphic. Indeed, any two lines in $ \mathbb {P} _{k}^{2} $ intersect, but the lines in $ X $ defined by $ w=x=0 $ and $ y=z=0 $ cannot intersect since their intersection would have all coordinates zero.

My question is about why the text's argument about "lines" works. I am confused because being a "line" is a property of the embedding in projective space, not of the abstract variety itself: an isomorphism need not send a line in $\mathbf P^2$ to a line in $X$.

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Tracing through the proof, one gets an explicit birational equivalence \[ \phi:\mathbf P^2 \dashrightarrow X,\qquad [a:b:c]\longmapsto [c^2:ac:bc:ab]. \] Indeed, \[ (ac)(bc)-(c^2)(ab)=0, \] so the image lies in $X$. Its inverse on the open set $w\neq 0\subset X$ is \[ \psi:X\dashrightarrow \mathbf P^2,\qquad [w:x:y:z]\longmapsto [x:y:w]. \]