This started as a competing answer, but now it is just a computation of what Donu has stated already. It might still be useful for some.

First let's introduce some notation: $\pi:Y\to X$ is the double cover and $Z'=(\pi^*Z)_{\mathrm{red}}$ is the reduced pre-image of $Z$.

1

In the situation of the question we have that $$ \pi_*\Omega_Y^p(\log Z')\simeq \Omega_X^p(\log Z) \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$} $$ where $\mathscr L$ is (as Donu already said) the anti-invariant part of the direct image of $\mathscr O_Y$ to $X$ under the natural $\mathbb{Z}/2$ action.

Since $\pi$ is finite, all higher direct images vanish and hence we have a similar isomorphism for cohomology: $$ H^q(Y,\Omega_Y^p(\log Z'))\simeq H^q(X, \pi_*\Omega_Y^p(\log Z'))\simeq H^q(X, \Omega_X^p(\log Z) )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1}) $$ by (3.22) of Esnault-Viehweg, Lectures on Vanishing Theorems.

2

If one is interested in Hodge numbers of the open manifolds $X\setminus Z$ and $Y\setminus Z'$, then this should be good. Otherwise we need to connect these to the non-logarithmic sheaves. For that probably the best tool is the following short exact sequence:

$$ 0 \to \Omega_X^p \to \Omega_X^p(\log Z)\to \Omega_Z^{p-1} \to 0. $$

(The existence of this short exact sequence is a simple exercise, or can be found in (2.3) of ibid.

There is of course an equivalent one on $Y$ with $Z'$:

$$ 0 \to \Omega_Y^p \to \Omega_Y^p(\log Z')\to \Omega_{Z'}^{p-1} \to 0. $$

Aha!

Until this point I thought that I was going to get a different answer than Donu and that was the main reason I even started writing, but now it seems that I might get from this what Donu stated.

The point is, $\pi$ induces an isomorphism $Z'\to Z$ and hence the right hand side of the two short exact sequences are the same. So if we add $\Omega_X^p(\log Z) \otimes \mathscr L^{-1}$ to the first short exact sequence and push-forward the second short exact sequence, then we get $$ 0 \to \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \to \Omega_X^p(\log Z)\oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big)\to \pi_*\Omega_{Z}^{p-1} \to 0. $$$$ 0 \to \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \to \Omega_X^p(\log Z)\oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big)\to \Omega_{Z}^{p-1} \to 0. $$ and

$$ 0 \to \pi_*\Omega_Y^p \to \pi_*\Omega_Y^p(\log Z')\to \pi_*\Omega_{Z'}^{p-1} \to 0. $$

Now, since $\pi|_{Z'}:Z'\to Z$ is an isomorphism and by $(\star)$ we get that these two short exact sequences are the same, so we have

$$ \pi_*\Omega_Y^p\simeq \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$} $$

and we get

$$ H^q(Y,\Omega_Y^p)\simeq H^q(X, \pi_*\Omega_Y^p)\simeq H^q(X, \Omega_X^p )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1}) $$ as stated by Donu.