The Hodge  numbers of a covering Let $X$ be a Kahler manifold and $Z\subset X$ 
be a smooth hypersurface. How to compute the Hodge 
diamond of the double covering $Y\to X$ ramified 
over $Z$? (And what I have to know? Would the map 
$H^*(X)\to H^*(Z)$ be enough?) 
P.S. I tried the Gysin sequence, but it looks 
like there are many loose ends.
 A: The answer goes back to Esnault and Viehweg:
$$H^q(Y,\Omega_Y^p)= H^q(X,\Omega_X^p)\oplus H^q(X,\Omega_X^p(\log Z)\otimes L^{-1})$$
where $L$ is the anti-invariant part of the direct image of $O_Y$ to $X$ under the
natural $\mathbb{Z}/2$ action.
A: This is more like a comment that in order to get a positive answer you need to specify more data.
Let us consider the simplest situation when $Z$ is empty, so $H^{\ast}(X)\to H^{\ast}(Z)$ is 
the zero map. It might happen in this situation, that there is more than one (unramified) double cover of $X$ and moreover double covers have different Betti numbers. Of course Hodge diamonds will be different as well.
Example. Consider the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Let $X=(E\times E)/\sigma$ be the quotient of $E\times E$ by the following fixed point free involution: $(x,y)\to (-x,y+\frac{1}{2})$. Then $X$ has several unramified double covers. One obvious cover is $E\times E$ that we started with. A different cover of $X$ can be obtained from $E\times E$ by taking a double cover of the first factor and then taking a quotient by the lift of $\sigma$. This cover has same topology as $X$ and hence different $b_1$ from $E\times E$. Indeed, $b_1(X)=2\ne 4=b_1(E\times E)$.
A: 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
$$
0 \to \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \to \Omega_X^p(\log Z)\oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big)\to \Omega_{Z}^{p-1} \to 0.
$$
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.
