Let $P$ be an acyclic object of $Ch_R$, let $P^{(k)}$ be the chain subcomplex of $P$ which agrees with $P$ above the degree $k-1$, contains $Bd_{k-1}P$ in degree $k-1$, and vanishes below degree $k-1$. Then the acyclicity condition gives isomorphisms $P^{(k)}/P^{(k+1)}\cong D_k(Cy_{k-1}P)$
NOTATION: $Bd_{k-1}P := \partial P_k$ and $Cy_{k-1}:= \ker \partial \colon B_{k-1} \to B_{k-2}$.
$D_n(A)$ is the chain complex all zero except in degree $n$ and $n-1$ where it is $A$ (the non trivial differential is the identity).
I have trouble proving the bolded claim: I interpreted the quotient as a degree wise quotient, and in fact I obtain $0$ in degree $\leq k-1$ and $\geq k+1$, in degree $k-1$ I obtain $Bd_{k-1}P = Cy_{k-1}P$ (by acyclicity), but in degree $k$ I obtain $P_k /Bd_{k}P$, which is not isomorphic to $Cy_{k-1}P$
I encountered this passage in a proof of a more general lemma, where by hypothesis we have that each $P_k$ is a projective module, I don't know if this could help, because such hypothesis is used later in the proof.
Maybe it's worth noting that you obtain a chain complex $$ \rightarrow 0 \rightarrow P_k/Bd_{k}P \rightarrow Bd_{k-1}P \rightarrow 0 \rightarrow $$
So if we prove that the above sequence is indeed exact, we are done (using the acyclicity condition ($Bd_{k-1}P=Cy_{k-1}P$ in your notation)
But this is indeed the case. The boundary map (the map induced by it actually) is injective because you are quotienting its kernel (which are exactly the cycle there - recall acyclicity-) and it is surjective because on the right you have it's image. hence you have an isomorphism and by acclivity you obtain your "disk"