Can anyone explain me what is the homomorphism $\mathcal{O}_X(V)\longrightarrow\mathcal{O}_X(U)$ such that $\alpha $ is integral over $\mathcal{O}_X(V)$?
Notice that foe Qing Liu a normal scheme is a scheme which is irreducible and normal at all its points.
There is no such ring homomorphism. There is a ring homomorphism in the other direction; this is just the restriction map.
Normal schemes are integral. If $\xi$ denotes the generic point of $X$, then for all open subsets $U$ of $X$, the ring of sections over $U$, denoted $\Gamma(U, X) = \mathcal O_X(U)$, can be identified with a subring of $K(X) = \mathcal O_\xi$. In fact, we can identify all local rings with subrings of $K(X)$, so that $\mathcal O_X(U) = \bigcap_{x\in U} \mathcal O_x$ holds. So the restriction map $\mathcal O_X(U)\to \mathcal O_X(V)$ is just the inclusion of these subrings. It is also true that in case such an open subset of $X$, say $V$, is an open affine subset of $X$, then $K(X)$ is the fraction field of $\mathcal O_X(V)$
In the proof above, $\alpha$ is regarded as an element of $K(X)$ under the inclusion $\mathrm{Frac}(\mathcal O_X(U))\to K(X) = \mathrm{Frac}(\mathcal O_X(V))$.