Let  $n=2$ and $k=3$, and suppose by the sake of simplicity that the three lines are in general position. Then, up to projective transformations, we can assume that they are the three coordinate lines $\ell_1$, $\ell_2$, $\ell_3$ given by $z_0=0$, $z_1=0$, $z_2=0$, respectively.

Then your function field is simply $\mathbb{C}(x, \, y)(\sqrt{x}, \, \sqrt{y})$, where $x=z_1/z_0$, $y=z_2/z_0$, and the affine equation of your $(\mathbb{Z}/2\mathbb{Z})^2$-cover $X \to \mathbb{P}^2$ on the chart $z_0 \neq 0$ is $$(x, \, y) \mapsto (x^2, \, y^2).$$

Note that $X$ is projective, since it is a finite cover of a projective variety; in fact, you can cover $X$ with three of these affine charts, corresponding to the three standard charts for $\mathbb{P}^2$. 

A moment of thought shows that $X = \mathbb{P}^2$, and that the *global* equation of your bi-double cover is $$\pi \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad (z_0, \, z_1, \, z_2) \mapsto (z_0^2, \, z_1^2, \, z_2^2).$$

It is an instructive exercise to factor $\pi$ through the three *singular* double covers $$X_i \to \mathbb{P}^2, \quad i=1,\, 2, \, 3$$ corresponding to the three involutions in the Klein group $(\mathbb{Z}/2\mathbb{Z})^2$.