References to understand $K3$ surface as a double cover of $\mathbb{P}^2$ ramified along a sextic My goal is to understand that 

$2:1$ cover of $\mathbb{P}^2(\mathbb{C})$ ramified along a sextic is a $K3$ surface.

My main problem is in understanding the theory of ramified covering of $\mathbb{P}^2.$ Since $\mathbb{P}^2$ is compact, there are no global non constant holomorphic functions there. So one considers the section in a line bundle over $\mathbb{P}^2.$ I don't understand what happens after this. Is there a nice reference to understand the ramified covering of projective space? 
Just as a matter of fact, I know about adjunction formula, holomorphic line bundles, divisors...(and of course the defintion of K3 surfaces.) 
 A: I don't know a specific reference but I can give an explanation of how you classify ramified double covers of any variety $X$ over an algebraically closed field. 
Say $f :Y \to X$ is a ramified double cover. Then $f$ is finite of degree two so we can consider $\mathcal{A}:= f_*\mathcal{O}_Y$ which is a rank two locally free sheaf on $X$. There is the natural map $j:\mathcal{O}_X \to \mathcal{A}$ making $\mathcal{A}$ an $\mathcal{O}_X$ algebra. On the other hand, there is a norm map $T_{Y/X} : \mathcal{A} \to \mathcal{O}_X$ which on affines is just the usual trace map for an integral extension of ring. Checking on affines, we get that $T_{Y/X}\circ j : \mathcal{O}_X \to \mathcal{O}_X$ is multiplication by $2$ (the degree of the integral ring extension). Therefore $\frac{1}{2} T_{Y/X}$ provides a splitting and so we get that 
$$
\mathcal{A} = \mathcal{O}_X \oplus \mathcal{L}
$$
where $\mathcal{L}$ is some line bundle. 
Since $\mathcal{A}$ splits like this, the multiplication $m : \mathcal{A} \otimes \mathcal{A} \to \mathcal{A}$ decomposes as maps 
$$
\mathcal{O}_X \otimes \mathcal{O}_X \to \mathcal{A} \enspace \enspace \enspace \mathcal{O}_X \otimes \mathcal{L} \to \mathcal{A} \enspace \enspace \enspace \mathcal{L} \otimes \mathcal{O}_X \to \mathcal{A} \enspace \enspace \enspace \mathcal{L} \otimes \mathcal{L} \to \mathcal{A}.
$$
The first three maps are uniquely determined by the fact that $\mathcal{A}$ is an  $\mathcal{O}_X$ algebra and so multiplication in $\mathcal{A}$ by $\mathcal{O}_X \subset \mathcal{A}$ is just the usual multiplication by $\mathcal{O}_X$ on an $\mathcal{O}_X$ module. 
Thus all the information of the algebra structure of $\mathcal{A}$ is contained in the map $\mathcal{L} \otimes \mathcal{L} \to \mathcal{A} = \mathcal{O}_X \oplus \mathcal{L}$. We can split this up further into maps $\mathcal{L} \otimes \mathcal{L} \to \mathcal{O}_X$ and $\mathcal{L} \otimes \mathcal{L} \to \mathcal{L}$. 
Now let us look at stalks. Since this is a double cover, the stalk of $\mathcal{A}$ at $x \in X$ is either $k^2$ or $k[\epsilon]/(\epsilon^2)$ with the latter happening if and only if $f$ is ramified above $x$. Then a stalkwise computation shows that the map $\mathcal{L} \otimes \mathcal{L} \to \mathcal{L}$ vanishes so that we only have a map $\mathcal{L} \otimes \mathcal{L} \to \mathcal{O}_X$ and this map is zero precisely when the stalk is $k[\epsilon]/(\epsilon^2)$, i.e. over the branch locus. 
In this way, to any double cover $f: Y \to X$ we can associate a line bundle $\mathcal{L}$ and a map $\mathcal{L}^{2} \to \mathcal{O}_X$ that vanishes exactly on the branch locus of $f$. Dualizing we get a map $\mathcal{O}_X \to \mathcal{L}^{-2}$, that is, a global section of $\mathcal{L}^{-2}$ whose vanishing locus is the branch locus of $f$. One can show that this gives a one to one correspondence between branched double covers and pairs of a line bundle $\mathcal{L}$ and a section of $\mathcal{L}^{-2}$. 
$\textbf{EDIT}$ What I wrote about how to go backwards and construct a double cover of $X$ given a section of $\mathcal{L}^{-2}$ was wrong. I will update when I figure out a nice way to do this. 
A: Dori Bejleri's answer gives an excellent general overview of double cover constructions. Let's work out some of the details in the case you want to understand. (Sorry if this is long-winded; I don't know the textbook treatment, so instead you get my stream of consciousness.)
First of all, suppose that $f:Y \rightarrow X$ is any generically finite morphism of smooth varieties (in characteristic 0, say). Then pullback of top-degree differential forms gives a map of sheaves $$f^*: \omega_X \rightarrow \omega_Y$$
which you can check is injective. So $f^* \omega_X$ is a subsheaf of $\omega_Y$. We want to understand what the "difference" is.
At this point, let's specialise to a double cover $f: Y \rightarrow X$ of complex surfaces. As we know from Dori's answer, this is determined by a section $s$ of a certain bundle $L^{-2}$ on $X$. To understand this in a more down-to-earth way, let's focus on an open set $U \subset X$ in which $L$ is trivialised: then on $U$ we can identify $s$ with a regular function, and the open set $f^{-1}(U) \subset Y$ is exactly the subset $\{(x,y) \in U \times \mathbf A^1 \mid y^2=s(x) \}$.    
(For the statement you want to be strictly true, one should assume that the ramification locus is a smooth sextic, so from now on I assume that $s(x)=0$ is a smooth curve in $X$.)
Fix a point $p=(x,y)$ of $Y$. For simplicity let's work in the classical (or analytic) topology. If $s(x) \neq 0$ then by the implicit function theorem and the equational description of $Y$ above, there's a (classical) open neighbourhood $V$ of $p$ in which the map $f:(x,y) \rightarrow x$ is an isomorphism from $V$ onto an open set in $X$. In this case we certainly have $(f^* \omega_X)_{|V} = (\omega_Y)_{|V}$.
On the other hand suppose that we look at a point $p=(x,0)$, that is, a point where $s(x)=0$. Then we can choose coordinates $(u,v)$ locally near $p$ on $Y$, and $(z,w)$ locally near $f(p)$ on $X$, in which the map $f$ is given by the formula $f: (u,v) \mapsto (u,v^2)$. A local generator near $f(p)$ of the sheaf $\omega_X$ of 2-forms is given by $\omega = dz \wedge dw$: this pulls back to $f^* \omega= du \wedge d(v^2) = -2v (du \wedge dv)$. (Note that in these coordinates the ramification divisor on $Y$ is given by the equation $v=0$.)  Now $du \wedge dv$ is a local generator of $\omega_Y$ near $p$, so in this open set we find that $f^* \omega_X = v \cdot \omega_Y$. 
Putting all this together into a global description, we find $f^* \omega_X = I_R \cdot \omega_Y$ where $I_R$ is the ideal sheaf of the ramification divisor on $Y$. One could also write this as $f^* \omega_X = O(-R) \otimes \omega_Y$, or again, in divisor notation, as $f^* K_X = K_Y - R$. 
Finally we observe that the branch divisor $B$ on $X$ pulls back to $f^*B = 2R$ (e.g. because its pullback is defined locally by the equation $v^2=0$ in the previous notation). So we can rewrite our last formula again as $f^*K_X = K_Y - \frac{1}{2} f^*B$, or more elegantly, $K_Y=f^*(K_X+\frac{1}{2}B)$.
In your example $X=\mathbf P^2$, so $K_X=3H$, and $B$ is a sextic; plugging these into the formula above gives $K_Y=0$. This explains why $Y$ has trivial canonical bundle. 
