This is just an expanded version of Tyler's comment, I think.

Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$, and let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.

I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$.

If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y_i\otimes z_i]$ where the coproduct of $x$ is $\Sigma y_i\otimes z_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get

$[a,bx]=[ab,x]=[a\otimes b,\Sigma y_i\otimes z_i]=\Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,\Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.

The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.

It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.

Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.

This is just an expanded version of Tyler's comment, I think.

Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$. Let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.

I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$.

If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y_i\otimes z_i]$ where the coproduct of $x$ is $\Sigma y_i\otimes z_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get

$[a,bx]=[ab,x]=[a\otimes b,\Sigma y_i\otimes z_i]=\Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,\Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.

The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.

It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.

Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.