The polynomial answer by David Speyer (also Theo Johnson-Freyd's comment) is really neat. Here is a generalisation of that technique, stated in the language of group actions and group cohomology.
Let $G$ and $A$ be groups (eventually $A$ will be abelian), and $\pi \colon X \to Y$ 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 $A$. Equivalently, $X$ is a set with a $(G \times A)$-action such that the $A$-action is free, and $Y = X/A$. If $A$ is abelian and $Y$ finite, we will use this data to construct a homomorphism $G \to A$.
Example. The example we're interested in is the following: let $S$ and $T$ be finite sets, and consider the action of $S(S) \times S(T)$ on $X = \operatorname{Inj}(S,T)$ by pre- and postcomposition. The action of $S(S)$ is free since injections are monic. Concretely, we want $\lvert S\rvert = 2$ and $T = \lvert n\rvert$ to get $S_2 \times S_n$ acting on $\operatorname{Inj}(2,T) \cong T^2 \setminus \Delta_T$ (pairs of distinct elements of $T$, as in David Speyer's answer). The $S_2$-action is free, so we will produce a homomorphism $S_n \to S_2$ that is easily seen to be equivalent to one in David Speyer's answer (take for $s \colon Y \to X$ the section that takes the equivalence class $\{(i,j),(j,i)\}$ to the representative with $i<j$).
Lemma. Let $G$, $A$, and $\pi \colon X \to Y$ be as above.
- If $s \colon Y \to X$ is a (set-theoretic) section, then there is a unique map $f = f_s \colon G \times Y \to A$ such that $f(g,y)gs(y) = s(gy)$ for all $g \in G$, $y \in Y$. It satisfies $f(gh,y) = f(g,hy)f(h,y)$ for all $g,h \in G$, $y\in Y$.
- If $t \colon Y \to X$ is another section, there exists a unique map $c \colon Y \to A$ such that $t(y)=c(y)s(y)$ for all $y \in Y$. It satisfies $f_t(g,y) = c(gy)f_s(g,y)c(y)^{-1}$ for all $g \in G$, $y \in Y$.
- If $A$ is abelian and $Y$ finite, then $f \colon G \to A$ given by $\prod_{y\in Y} f({-},y)$ is a group homomorphism independent of the section $s$.
This definitely has a "crossed homomorphism as $H^1(G,A)$" flavour, except in a relative setting. Below, we reinterpret this as a ‘norm’ or 'corestriction' map. (Admittedly, I myself have never been super happy about these types of elementary cocycle computations, as they themselves feel like magic. But maybe a more general type of magic than the other arguments.)
Proof of Lemma. For $g \in G$ and $y \in Y$, we have $\pi(gs(y)) = gy = \pi(s(gy))$ since $\pi$ is equivariant and $s$ is a section. The first statement in (1) then follows since $A$ acts simply transitively on the fibres of $\pi$. For the second, we compute $s(ghy)$ in two ways: $$f(gh,y)ghs(y) = s(ghy) = f(g,hy)gs(hy) = f(g,hy)gf(h,y)hs(y).$$ The statement follows since the actions of $G$ and $A$ commute and again using the torsor property.
The first statement in (2) is again clear from the torsor property. For the second, the defining properties of $f_s$, $f_t$, and $c$ give $$f_t(g,y)gc(y)s(y) = f_t(g,y)gt(y) = t(gy) = c(gy)s(gy) = c(gy)f_s(g,y)gs(y).$$ The result again follows from the torsor property.
The map $f \colon G \to A$ in (3) is a homomorphism by the second statement in (1), and independent of the section $s \colon Y \to X$ by the second statement in (2). $\square$
Remark. I am fairly certain that the construction above can be identified with corestriction, at least once we fix base points $x \in X$ and $y = \pi(x) \in Y$.
Assume for simplicity that the $G$-action on $Y$ is transitive, so that $Y \cong G/H$ with $H = \operatorname{Stab}_G(y)$. Note that the map $\operatorname{Stab}_{G \times A}(x) \to H$ is an isomorphism, so we get a natural projection $\phi \colon H \to A$. The inclusion $\iota \colon H \to G$ induces a corestriction map $H^1(H,A) \to H^1(G,A)$, and the map $f \colon G \to A$ should be the corestriction of $\phi$ (I think it differs by a sign because of my choice of $f(g,y)$). I actually find my construction above more natural (as it depends on fewer choices); for instance we don't need to choose a base point, and we only need a section $Y \to X$ and not all the way $Y \cong G/H \to G$.
Example. In the example of $S_n$, picking $x = (1,2)$ and $y = \{1,2\}$ gives $$H = S_2 \times S_{n-2} \subseteq S_n,$$ with its natural projection $H \to S_2$. Clearly corestriction $H^1(H,\{\pm 1\}) \to H^1(S_n,\{\pm 1\})$ gives a homomorphism $S_n \to S_2$, but it's not very clear from this perspective that it takes $(12)$ (and therefore any transposition) to $-1$ (again, this computation requires a section $S_n/H \to S_n$, which is considerably more computationally involved than my construction above).
Rambling. It is conceivable that this construction can be resphrased in terms of the left Kan extension \begin{align*} \iota_! \colon H\text{-}\mathbf{Set} &\to G\text{-}\mathbf{Set} \\ Z &\mapsto (G \times Z)/H \cong \coprod_{[g] \in G/H} Z. \end{align*} It seems the finite index assumption only comes in once we're taking products in an abelian group, so it's still a little unclear what's going on here. Maybe reverse engineering Bjorn Poonen's excellent answer gives a more explicit way to do the corestriction without choosing a base point!