The thing that Hatcher does that is different from, say, Steenrod and Epstein as Charles mentioned, is that rather than construct anything on cohomology rings, he constructs an operation on Eilenberg-Maclane spaces. The $n$th cohomology group of any space $X$ with coefficients in $G$ is naturally isomorphic to the group of homotopy classes of maps from $X$ to an Eilenberg-Maclane space $K(G, n)$: this identification is nice and functorial because if you map $X\rightarrow Y$ you can map $X\rightarrow Y\rightarrow K(G, n)$ to get a pullback map on cohomology groups. Now if you want to make a cohomology operation, which can be defined naturally on cohomology groups for any $X$, all you need is an operation $K(G, n)\rightarrow K(H, m)$ and by composing with that you get a natural map $H^n(X; G)\rightarrow H^m(X;H)$.

So Hatcher's idea is to take a kind of (smash) product of $K(\mathbb Z/2, n)$ (which he calls $K_n$) with itself, and map that to $K_{2n}$ in a way that will be like the cup-product square in dimension $n$, since $Sq^n(\alpha)=\alpha^2$ when $\alpha$ is of dimeniosn $n$. More explicitly, since $H^n(X\wedge X)$ is isomorphic to $H^n(X)\otimes H^n(X)$ by the Kunneth formula, the point is to find an element of $H^{2n}(X\wedge X)$ which is the cup-product square. His notation is NOT standard.

Intuitively, the idea is to do this for $K_n$ so that it works on all $X$, and to put the copies of $K_n$ back together by quotienting by a $\mathbb Z/2$ action. But the $\mathbb Z/2$-action on $K_n\wedge K_n$ given by switching is not free, so to make it free he takes $S^\infty\times K_n\wedge K_n$ and uses the antipodal map on $S^\infty$ to map $(x, y, z)$ to $(-x, z, y)$ and give a free $\mathbb Z/2$ action. (The coordinates here are really in $S^\infty\times K_n\times K_n$). Then the whole rest of the construction is about dealing with the extra things that $\mathbb RP^\infty$ gives you, and from there he can calculate what happens to homology elements of $\wedge X$ when it is mapped to $K_{2n}$. That tells you the action of $Sq^{n-i}$ on $H^i(X)$.

Hope that helps.

1

The thing that Hatcher does that is different from, say, Steenrod and Epstein as Charles mentioned, is that rather than construct anything on cohomology rings, he constructs an operation on Eilenberg-Maclane spaces. The $n$th cohomology group of any space $X$ with coefficients in $G$ is naturally isomorphic to the group of homotopy classes of maps from $X$ to an Eilenberg-Maclane space $K(G, n)$: this identification is nice and functorial because if you map $X\rightarrow Y$ you can map $X\rightarrow Y\rightarrow K(G, n)$ to get a pullback map on cohomology groups. Now if you want to make a cohomology operation, which can be defined naturally on cohomology groups for any $X$, all you need is an operation $K(G, n)\rightarrow K(H, m)$ and by composing with that you get a natural map $H^n(X; G)\rightarrow H^m(X;H)$.
So Hatcher's idea is to take a kind of (smash) product of $K(\mathbb Z/2, n)$ (which he calls $K_n$) with itself, and map that to $K_{2n}$ in a way that will be like the cup-product square in dimension $n$, since $Sq^n(\alpha)=\alpha^2$ when $\alpha$ is of dimeniosn $n$. More explicitly, since $H^n(X\wedge X)$ is isomorphic to $H^n(X)\otimes H^n(X)$ by the Kunneth formula, the point is to find an element of $H^{2n}(X\wedge X)$ which is the cup-product square. His notation is NOT standard.
Intuitively, the idea is to do this for $K_n$ so that it works on all $X$, and to put the copies of $K_n$ back together by quotienting by a $\mathbb Z/2$ action. But the $\mathbb Z/2$-action on $K_n\wedge K_n$ given by switching is not free, so to make it free he takes $S^\infty\times K_n\wedge K_n$ and uses the antipodal map on $S^\infty$ to map $(x, y, z)$ to $(-x, z, y)$ and give a free $\mathbb Z/2$ action. (The coordinates here are really in $S^\infty\times K_n\times K_n$). Then the whole rest of the construction is about dealing with the extra things that $\mathbb RP^\infty$ gives you, and from there he can calculate what happens to homology elements of $\wedge X$ when it is mapped to $K_{2n}$. That tells you the action of $Sq^{n-i}$ on $H^i(X)$.