This answer grew out of an attempt to understand the polynomial product formula in David Speyer's answer (also Theo Johnson-Freyd's comment). It has been superseded by Bjorn Poonen's excellent self-contained answer.
If $T$ is a totally ordered set of $n$ elements, the product in Speyer's answer is defined in terms of the section to $$T^2 \setminus \Delta_T \to {T \choose 2}$$ taking $S \subseteq T$ to $(\min(S),\max(S))$, but clearly does not depend on the choice of this section (swapping a pair of indices induces a $-1$ on the top and bottom). This is reminiscent of a norm or corestriction in group cohomology. An earlier version of my answer worked this out in a more general setting using cocycles and crossed homomorphisms, but thanks to Poonen's answer and Benjamin Steinberg's comment, I can now do this without using cocycles.
If $S$ and $T$ are finite sets, then $S(T) \times S(S)$ acts on $X = \operatorname{Inj}(S,T)$ by post- and precomposition. The action of $S(S)$ is free since injections are monic, and we denote the quotient $X/S(S)$ by $Y={T \choose S}$. Concretely, we want $\lvert S\rvert = 2$ and $\lvert T \rvert = n$ to get $S_n \times S_2$ acting on $\operatorname{Inj}(2,T) \cong T^2 \setminus \Delta_T$. This is an example of the following more general object:
Setup. Let $G$ and $A$ be groups (eventually $A$ will be abelian), let $X$ be a set with a $(G \times A)$-action such that the $A$-action is free, and set $Y = X/A$. Equivalently, $\pi \colon X \to Y$ is a $G$-equivariant $A$-torsor, in the sense that $\pi$ is a $G$-equivariant map of $G$-sets such that \begin{align*} \mu \colon A \times X &\to X \underset Y\times X \\ (a,x) &\mapsto (ax,x) \end{align*} is a $G$-equivariant bijection, where $G$ acts trivially on $A$. In other words, this is an $\underline{A}$-torsor on the slice category $G\text{-}\mathbf{Set}/Y$, where $\underline{A}$ is the 'constant object' on $A$ (the pullback of $A$ from the terminal Grothendieck topos $\mathbf{Set}$).
The internal hom $\mathbf{Hom}_Y(Y,{-}) \colon G\text{-}\mathbf{Set}/Y \to G\text{-}\mathbf{Set}$ preserves limits, so takes the above to the $A^Y$-torsor $\mathbf{Hom}_Y(Y,X)$ of sections to $\pi$ (this is the torsor in Poonen's answer). Here, $A^Y$ is no longer a constant object: it has a natural $G$-action. Such a torsor corresponds to an element $\phi \in H^1(G,A^Y)$ (nonabelian cohomology, although we will now assume $A$ is abelian), i.e. a crossed homomorphism $\phi \colon G \to A^Y$. Given a section $s \colon Y \to X$ of $\pi$, the crossed homomorphism $\phi \colon G \to A^Y$ can be computed via the composition $$\begin{array}{ccccccc}G & \to & X \underset Y\times X & \stackrel\sim\leftarrow & A \times X & \stackrel{\pi_1}\to & A \\ g & \mapsto & (gs(y),s(gy)).\! & & & & \end{array}$$ In other words, $\phi(g)(y)$ is the unique $a \in A$ such that $gs(y) = as(gy)$. Different choices of section give crossed homomorphisms that differ by a principal crossed homomorphism.
If $A$ is abelian and $Y$ finite, then there is a ($G$-equivariant) multiplication map $\Pi \colon A^Y \to A$, giving a map $H^1(\Pi) \colon H^1(G,A^Y) \to H^1(G,A)$. Since the $G$-action on $A$ is trivial, this is just $\operatorname{Hom}(G,A)$, so $H^1(\Pi)(\phi)$ is a homomorphism $G \to A$.
In the case of $S_n \times S_2$ $\style{display: inline-block; transform: rotate(90deg)}{\circlearrowright}$ $\!\operatorname{Inj}(2,n)$, this gives a homomorphism $S_n \to S_2$. Given a section $s \colon {T \choose 2} \to T^2 \setminus \Delta_T$, write $s_1, s_2 \colon {T \choose 2} \to T$ for the projections $\pi_i \circ s$. Then the crossed homomorphism $\phi \colon S_n \to \{\pm 1\}^{T \choose 2}$ can be identified with $$\phi(\sigma)(S) = \frac{x_{\sigma(s_1(S))}-x_{\sigma(s_2(S))}}{x_{s_1(\sigma(S))}-x_{s_2(\sigma(S))}}.$$ Taking the product over all $S \in {T \choose 2}$ recovers the formula in Speyer's answer when $s$ is given by $S \mapsto (\min(S),\max(S))$.
Remark. One thing I still find weird about this is that it almost works to give a homomorphism $S_n \to S_3$ as well, by taking $\operatorname{Inj}(3,n)$ instead of $\operatorname{Inj}(2,n)$. It somehow only fails because $S_3$ is not abelian.
Relation to corestriction.
If the $G$-action on $Y$ is transitive, then choosing a point $y \in Y$ gives an isomorphism $G/H \stackrel\sim\to Y$ of $G$-sets, where $H = \operatorname{Stab}_G(y)$. Left Kan extension \begin{align*} H\text{-}\mathbf{Set} &\to G\text{-}\mathbf{Set} \\ Z &\mapsto (G \times Z)/H \cong \coprod_{G/H} Z \end{align*} identifies $H\text{-}\mathbf{Set}$ with the slice topos $G\text{-}\mathbf{Set}/Y$, and $A^Y$ is the coinduction of the trivial $H$-module $A$. For $\lvert Y\rvert = [G:H]$ finite, this is also the induction, so Shapiro's lemma says that $$H^1(G,A^Y) \cong H^1(H,A) = \operatorname{Hom}(H,A).$$ The map $H^1(H,A) \to H^1(G,A)$ in this case is called corestriction.
The homomorphism $H \to A$ can be deduced immediately from the Setup: if $x \in X$ is a lift of $y$, then $$\operatorname{Stab}_{G \times A}(x) \to \operatorname{Stab}_G(y) = H$$ is an isomorphism, so its inverse gives a projection $H \to A$. In our main example, the point $\{1,2\} \in {T \choose 2}$ has stabiliser $H = S_2 \times S_{n-2} \subseteq S_n$, and the above shows that the sign $S_n \to S_2$ is the corestriction of the projection $H \to S_2$.