Here's a somewhat more elementary argument that (S2) implies the Hartogs condition. More precisely, I will show that if $X$ is an (S2) noetherian scheme, then any rational function defined outside a closed subset of codimension two can be extended to the whole domain. (This extension is unique by definition of a rational function.)

Assume, by way of contradiction, that $X$ is an (S2) noetherian scheme and $f$ is a rational function on $X$ that is defined outside a closed set of codimension at least two, but cannot be extended to the whole domain. Let $\mathscr{I}$ be the ideal of denominators of $f$; in other words, over an open affine $\operatorname{Spec} A$, $$I = \{g \in A \mid g f \in A\}.$$ This is well-defined as a sheaf since the ideal of denominators is preserved under flat pullback (and in particular, localization); see this mathoverflow question.

If $g \in A$ is a nonzerodivisor, then $g \in I$ if and only if $f = a / g$ for some $a \in A$, hence the name "ideal of denominators." One can check that the closed subscheme $Z \subset X$ corresponding to $\mathscr{I}$ is, set-theoretically, the "indeterminacy locus of $f$": the smallest closed subset such that $f$ is defined over $X \smallsetminus Z$. By hypothesis, $f$ can be defined outside a closed subset of codimension two, so $\operatorname{codim} Z \geq 2$. Equivalently, whenever $W$ is an irreducible component of $Z$, then the local ring $\mathscr{O}_{X,W}$ has dimension at least two. Since $X$ is assumed to be (S2), every maximal regular sequence in $\mathscr{O}_{X,W}$ has length at least two.

Since $W$ is an irreducible component of the subscheme corresponding to $\mathscr{I}$, it follows that the radical of $\mathscr{I}_W \subset \mathscr{O}_{X,W}$ is precisely the maximal ideal $\mathfrak{p}$. (Algebraically, $\mathfrak{p}$ is a minimal prime over $I$, and corresponds to the generic point of $W$.) Let $g,h \in \mathfrak{p}$ form a regular sequence (which exists since $X$ is (S2)). Replacing $g$ and $h$ by appropriate powers, we may assume that they are both contained in $\mathscr{I}_W$. By definition of regular sequence, $g$ is a nonzerodivisor. Since $h,g$ is a also a regular sequence, $h$ is a nonzerodivisor. Thus, $g$ and $h$, being nonzerodivisors that lie in the ideal of denominators, are in fact denominators of $f$: there exist $a, b \in \mathscr{O}_X,W$ such that $$\frac{a}{g} = \frac{b}{h} = f$$ $$ah = bg.$$ Since $g,h$ is a regular sequence, $h$ is a nonzerodivisor mod $g$. When we mod out by $g$, the equation above becomes $ah \equiv 0$, which would imply $a \equiv 0 \pmod{g}$. In other words, $a \in (g)$. But since $f = a/g$, this would imply that $f \in \mathscr{O}_{X,W}$, a contradiction since $f$ cannot be extended over $W$.

Here's a somewhat more elementary argument that (S2) implies the Hartogs condition. More precisely, I will show that if $X$ is an (S2) noetherian scheme, then any rational function defined outside a closed subset of codimension two can be extended to the whole domain. (This extension is unique by definition of a rational function.)

Assume, by way of contradiction, that $X$ is an (S2) noetherian scheme and $f$ is a rational function on $X$ that is defined outside a closed set of codimension at least two, but cannot be extended to the whole domain. Let $\mathscr{I}$ be the ideal of denominators of $f$; in other words, over an open affine $\operatorname{Spec} A$, $$I = \{g \in A \mid g f \in A\}.$$ This is well-defined as a sheaf since the ideal of denominators is preserved under flat pullback (and in particular, localization); see this mathoverflow question.

If $g \in A$ is a nonzerodivisor, then $g \in I$ if and only if $f = a / g$ for some $a \in A$, hence the name "ideal of denominators." One can check that the closed subscheme $Z \subset X$ corresponding to $\mathscr{I}$ is, set-theoretically, the "indeterminacy locus of $f$": the smallest closed subset such that $f$ is defined over $X \smallsetminus Z$. By hypothesis, $f$ can be defined outside a closed subset of codimension two, so $\operatorname{codim} Z \geq 2$. Equivalently, whenever $W$ is an irreducible component of $Z$, then the local ring $\mathscr{O}_{X,W}$ has dimension at least two. Since $X$ is assumed to be (S2), every maximal regular sequence in $\mathscr{O}_{X,W}$ has length at least two.

Since $W$ is an irreducible component of the subscheme corresponding to $\mathscr{I}$, it follows that the radical of $\mathscr{I}_W \subset \mathscr{O}_{X,W}$ is precisely the maximal ideal $\mathfrak{p}$. (Algebraically, $\mathfrak{p}$ is a minimal prime over $I$, and corresponds to the generic point of $W$.) Let $g,h \in \mathfrak{p}$ form a regular sequence (which exists since $X$ is (S2)). Replacing $g$ and $h$ by appropriate powers, we may assume that they are both contained in $\mathscr{I}_W$. By definition of regular sequence, $g$ is a nonzerodivisor. Since $h,g$ is a also a regular sequence, $h$ is a nonzerodivisor. Thus, $g$ and $h$, being nonzerodivisors that lie in the ideal of denominators, are in fact denominators of $f$: there exist $a, b \in \mathscr{O}_X,W$ such that $$\frac{a}{g} = \frac{b}{h} = f$$ $$ah = bg.$$ Since $g,h$ is a regular sequence, $h$ is a nonzerodivisor mod $g$. When we mod out by $g$, the equation above becomes $ah \equiv 0$, which would imply $a \equiv 0 \pmod{g}$. In other words, $a \in (g)$. But since $f = a/g$, this would imply that $f \in \mathscr{O}_{X,W}$, a contradiction since $f$ cannot be extended over $W$.