I will make use of the surjection operad to produce an explicit example of a (functorial) cochain whose boundary is the Adem-Relation. Let us show that $Sq^1Sq^1([x])=0$ for a cohomology class $x$ in degree 1.
Thus we have on cochains
$$(x\cup x)\cup_1(x\cup x)=\langle 121\rangle (\langle 12\rangle(x^{\otimes 2}))=(\langle 13412\rangle +\langle 12342\rangle)(x^{\otimes 4}).$$
The boundary of $\langle 123412\rangle\in Hom ((C^*)^{\otimes 4},C^*)$ is
$$\langle 23412 \rangle +\langle 13412 \rangle+\langle 12342 \rangle+\langle 12341 \rangle.$$
If we plug in the cycle $x^{\otimes 4}$, we can use that relabeling the numbers $1,..,4$ corresponds to swapping the indices and thus does not affect the result. Thus we have
$$\langle 12341\rangle(x^{\otimes 4})=\langle 23412\rangle(x^{\otimes 4}).$$
So if we plug in $x^{\otimes 4}$, two summands cancel out and the boundary is just the Adem-relation on cochain level.

Now it remains to find a simplicial set $X$ and $1$-cocycle $c$ such that $\langle 123412\rangle(c^{\otimes 4})\neq 0$, i.e. there is a two simplex on which the cochain $\langle 123412\rangle(c^{\otimes 4})$ does not vanish.

This happens for the $1$-cochain on the $2$-simplex $012$ given by
$c(01)=1,c(12)=1,c(02)=0$.