Is it possible to realize $K3$ surface as a ramified double cover of rational elliptic surface? If so, is there way to see an elliptic fibration structure on $K3$ from such cover? It seems to me one can use the divisor $6H - 2E\_{1} - \cdots - 2E\_{9}$ and the class $3H - E\_{1} - \cdots - E\_{9}$ in two fold cover gives the fiber class in $K3$? 



Question on K3 Surface
I slightly disagree with Tony's answer: 


If you want to obtain a smooth K3 surface you need to pick a ramification divisor that is smooth and is linear equivalent to twice the anti-canonical class. I.e., you have to pick two smooth fibers.


You can easily see this in terms of Weierstrass equation. Let $S$ be a rational elliptic surface. Since the Picard group of $\mathbb{P}^1$ is trivial you can pick a global minimal Weierstrass model for $S$, i.e., you there exists forms $A(s,t)$ and $B(s,t)$ on $\mathbb{P}^1$ of degree 4, resp., 6. such that $S$ is birational to the solution set of $y^2=x^3+Ax+B$ in some appropriate $\mathbb{P}^2$-bundle. If $S$ is sufficiently general then $S$ is actually isomorphic to its Weierstrass model. 


If we pick now two further forms $s(u,v)$ and $t(u,v)$ of degree 2 then
$y^2=x^3+A(s(u,v),t(u,v))x+B(s(u,v),t(u,v))$ is the Weierstrass model of an elliptic surface. For a general choice of the forms $s(u,v),t(u,v)$ this defines a smooth K3 surface.


