I am trying to show that $d²=0$ where $d$ is the derivation on $T(s^{-1}\bar{C})$ induced by the map $s^{-1}\bar{C}\to T(s^{-1}\bar{C})$ defined by $$s^{-1}x\mapsto -\sum (-1)^{|x_{(1)}|}s^{-1}x_{(1)}\otimes s^{-1}x_{(2)}$$ where $\bar{\Delta}(x)=\sum x_{(1)}\otimes x_{(2)}$.
I tried doing this by considering an element $s^{-1}c_1\otimes...\otimes s^{-1}c_k\in T(s^{-1}\bar{C})$. Because $d$ is linear and the elemets of $T(s^{-1}\bar{C})$ are sums of elements of this form this should suffice. Leibniz rule tells us that $$d(s^{-1}c_1\otimes...\otimes s^{-1}c_k)=\\ (\sum 1\otimes...\otimes 1 \otimes d\otimes 1\otimes ...\otimes 1)(s^{-1}c_1\otimes...\otimes s^{-1}c_k).$$ By writing out the formulas and keeping track of all the signs I managed to prove that when we apply $d$ twice, the term we get from first applying $d$ on the $i$'th position and then applying $d$ on the $j$'th position cancels with the term we get from first applying $d$ on the $j$'th position and then applying $d$ on the $i$'th position as long as $j\neq i,i+ 1$. However, I got stuck with the terms we get from first applying $d$ on the $i$'th position and then applying $d$ on the $(i+1)$'th position. I know coassociativity of $C$ should imply that these terms cancel with the terms we get by applying $d$ on the $i$'th position twice but I cannot prove it.
In the book Algebraic Operads by Loday and Vallette they write out a proof of it (page 44) but I do not understand it. They only consider $d^2$ applied to an element of the form $s^{-1}x$. Then I think that they identify terms of $$(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x):=\sum x_{(1)}\otimes x_{(2)}\otimes x_{(3)}$$ with terms of $$(1\otimes \bar{\Delta})\circ \bar{\Delta}(x):=\sum x_{(1)}'\otimes x_{(2)}'\otimes x_{(3)}'$$ and show how they cancel given the signs they get in $d²$. This is what I don't understand because coassociativity of $C$ only tells us $(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x)=(1\otimes \bar{\Delta})\circ \bar{\Delta}(x)$ but $$(\bar{\Delta}\otimes 1)\circ \bar{\Delta}(x)=\sum x_{(1)}\otimes x_{(2)}\otimes x_{(3)}$$ need not contain the same terms as $$(1\otimes \bar{\Delta})\circ \bar{\Delta}(x)=\sum x_{(1)}'\otimes x_{(2)}'\otimes x_{(3)}'.$$
Maybe someone who is familiar with the book and/or the theory can help bring clarity here=)
You are missing many signs from desuspension, which you did write in your first sentence, but didn't afterwards.
First, recall that since the differentail, call it $d$, induced on $\Omega C$, obtained from the comultiplication, is a derivation, and since $\Omega C$ is generated in degree $1$ by $V=s^{-1}\overline C$, it suffices you prove the claim on these generators. Now the differential is given as follows: take $s^{-1}c$ a generator, desuspend, and comultiply to get $c_{(1)}\otimes c_{(2)}$, and then multiply by $s^{-1}\otimes s^{-1}$ to get (from the Koszul rule and the interchange of $c_{(1)}$ and $s^{-1}$)
$$d(s^{-1}c) = -(-1)^{|c_{(1)}|}s^{-1}c_{(1)}\otimes s^{-1}c_{(2)}.$$
If you apply $d$ again, you will get two terms, one is
$$ (-1)^{|c_{(1)}|+|c_{(1)(1)}|}s^{-1}c_{(1)(1)}\otimes s^{-1}c_{(1)(2)}\otimes s^{-1}c_{(2)},$$
where the sign $|c_{(1)(1)}|$ comes from the same rule as above. The other is
$$ (-1)^{|c_{(1)}|+|c_{(2)(1)}|+|c_{(1)}|-1}s^{-1}c_{(1)}\otimes s^{-1}c_{(2)(1)}\otimes s^{-1}c_{(2)(2)},$$
where $|c_{(1)}|-1$ comes from $d$ going over $s^{-1}c_{(1)}$, so the overall sign is just $|c_{(2)(1)}|-1$.
To get the global sign of the first term, note that $|c_{(1)}|= |c_{(1)(1)}| +|c_{(1)(2)}|$, so the sign is $|c_{(1)(2)}|$. Writing everything in Sweedler notation, you get
$$(-1)^{|c_{(2)}|}c_{(1)}\otimes c_{(2)}\otimes c_{(3)} + (-1)^{|c_{(2)}|-1}c_{(1)}\otimes c_{(2)}\otimes c_{(3)} = 0$$