This is from Vakil's Foundations of Algebraic Geometry, 10.7.I
Let $X$ be an integral scheme with function field $K$. Let $L$ be an algebraic extension of $K$.
Then, the normalization of $X$ in $L$ is a scheme $v : \widetilde{X} \rightarrow X$ with $\widetilde{X}$ normal, integral and having function field $L$, $v$ is a dominant morphism that induces the inclusion $i : K \rightarrow L$ and is universal wrt this property. That is: for any scheme $f :Y \rightarrow X$ also satisfying the previous properties, there is a unique morphism $f' : Y \rightarrow \widetilde{X}$ such that $v \circ f' = f$.
If $X = Spec(R)$ is affine, then I expect $\widetilde{X}$ to be the integral closure of $R$ inside $L$, but this seems to not be the case as the then the factorization $v \circ f' = f$ is not unique, as it seems to depend on the choice of isomorphism $K(Y) \rightarrow L$.
For example, if $R = \mathbb{Z}, K = \mathbb{Q}, L = \mathbb{Q}[\sqrt{2}]$, then $\widetilde{R} = \mathbb{Z}[\sqrt{2}]$. If $Y$ is also $Spec(\mathbb{Z}[\sqrt{2}])$, then we get different morphisms $Y \rightarrow Spec(\widetilde{R})$ depending on if we take the isomorphism $K(Y) \simeq \mathbb{Q}[\sqrt{2}] \rightarrow L$ to be identity or conjugation. They correspond to identity or conjugation maps $\widetilde{R} = \mathbb{Z}[\sqrt{2}] \rightarrow O_Y(Y) = \mathbb{Z}[\sqrt{2}]$. Thus, the $f'$ is no longer unique. How do we fix this?
I think the only way to fix this problem is to choose a specific isomorphism $j : K(Y) \simeq L$, and require that the $f'$ induce $j$, so different $j$'s will result in different $f'$'s.
Yes. The isomorphism $L \to K(Y)$ belongs to the data of the normalization. This is meant with "having function field $L$" (similarly for $X$).
To be precise, we want a dominant $Y \to X$ together with a commutative diagram $$\require{AMScd} \begin{CD} K @>{\cong}>> K(X) \\ @VVV @VVV \\ L @>>{\cong}> K(Y) \end{CD}$$ The normalization is a terminal object in this category where $Y$ is normal.