Cap product Hochschild (co)homology

I will do the computation using bar symbols. (It is really hard from here to understand why the author in loc. cit. prefers the many tensors. Notations are important to fix the calculus.)

There are always sign conventions from textbook to textbook, here i am doing the computations barely without checking the reference.

We have for $f$ of degree one, and $g=m[a|b]$ the following computation using bar symbols: $$ \begin{aligned} g &= m[a|b]\ ,\\ f\frown g & =mf(a)[b]\ ,\\[2mm] d(f\frown g) &= -mf(a)d[b] \\ &= -mf(a)\; (b[]-[]b) \\ &= -mf(a)b[] +mf(a)[]b\\ &= -\color{blue}{mf(a)b[]} +\color{red}{bmf(a)[]}\\ &\qquad\text{and here is the sense for the bar symbols,}\\[2mm] \underbrace{df}_{\in C^2}\frown g &= df\frown m[a|b]\\ &= m\;df[a|b]\\ &= m\;fd[a|b]\\ &= m\;f\;(a[b]-[ab]+[a]b)\\ &= m\;(af(b)[]-f(ab)[]+\color{red}{f(a)[]b})\\[2mm] f\frown dg &= f\frown m\; d[a|b]\\ &= f\frown m\; (a[b]-[ab]+[a]b)\\ &= m\; (af(b)[]-f(ab)[]+\color{blue}{f(a)b[]})\\ \end{aligned} $$ Subtracting the two terms $df\frown g$ and $f\frown g$ from each other produces the cancellation of the unmarked terms, and the red/blue term in the result corresponds to the one in $d(f\frown g)$.

Here, we use the convention that scalars after the bar symbol, e.g. as in $mf(a)[]b$ can be "cyclically" moved in front (via the tensor product of that $A^e=A\otimes A^{\text{opp}}$, from the opp-part) to the $m\in M$, which has as structure of bimodule. (It is hard to do cyclic (co)homology without a notation supporting the cyclicity.)