<h3>Normality</h3>For an integral scheme, being normal (integrally closed in ones own fraction field) satisfies this property.  Indeed, suppose that $a, b \in A = \Gamma(X, O_X)$ are such that $a/b$ satisfy some polynomial $p(x) \in A[x]$.  Then $a|_U, b|_U$ satisfy the same polynomial after restriction to each (affine) set $U \subseteq X$.

Of course this shows up in many applications of things like Stein factorization.
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<h3>Semi-normality</h3>

A reduced ring $R$ is seminormal if for any finite extension $R \subseteq S$ satisfying the following two properties is an isomorphism.

 - The induced map on $\text{Spec}$'s is an isomorphism
 - The induced residue field extensions $k(r) \subseteq k(s)$ are isomorphisms for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.



The typical example of a seminormal ring is a node, the cusp $k[x^2,x^3]$ is not seminormal

Equivalently, $R$ is seminormal if and only if for any $a/b$ in the total ring of fractions of $R$, one has that if $(a/b)^2, (a/b)^3 \in R$ then $(a/b) \in R$ (see a paper by Swan, he might be assuming finitely many minimal primes, I forget the details).  It follows similarly that seminormality satisfies this property.

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<h3>Weak normality</h3>
Weak normality is similar to semi-normality.  A reduced ring is called weakly normal if for any finite *birational* extension $R \subseteq S$ satisfying the following properties is an isomorphism:

 - The induced map on $\text{Spec}$'s is an isomorphism
 - The induced residue field extensions $k(r) \subseteq k(s)$ are purely for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$. 

I do *NOT* know if weakly normal rings satisfy the sort of property asked for.  I do not think it is in the literature (but perhaps I am wrong).  I remember I convinced myself that they did not several years ago, but never wrote down an example carefully.