Fix a ring $R$; in the following, "module" will always mean "$R$-module". Chain complexes of modules are denoted as $C_\bullet$, and the differentials as $d$. For $k\in \mathbb N$, the complex $C_\bullet[k]$ has $C_n[k]:=C_{n+k}$, and all the differential multiplied by $(-1)^k$.
Let $A$ be a module, with a projective resolution $P_\bullet$, and let $B$ be a module with a projective resolution $Q_\bullet$. Let $\epsilon:P_0\to A$ and $\eta:Q_0\to B$ be the surjections of the respective projections. We can construct this bicomplex, call it $P\otimes Q$, which continues in all the first quadrant: $\require{AMScd}$ $$\begin{CD} P_0\otimes Q_2@< d\otimes Q_2<< P_1\otimes Q_2@<d\otimes Q_2 <<P_2\otimes Q_2 \\ @VVP_0\otimes dV @VV-P_1\otimes dV @VVP_2\otimes dV\\ P_0\otimes Q_1@< d\otimes Q_1<< P_1\otimes Q_1@< d\otimes Q_1<< P_2\otimes Q_1\\ @VVP_0\otimes dV @VV-P_1\otimes dV @VVP_2\otimes dV\\ P_0\otimes Q_0@< d\otimes Q_0<< P_1\otimes Q_0@< d\otimes Q_0<< P_2\otimes Q_0 \end{CD}$$
We have also this particular bicomplex in the first quadrant, call it $A\otimes Q$: $$\begin{CD} A\otimes Q_2@<<< 0@<<< 0 \\ @VVA\otimes dV @VVV @VVV\\ A\otimes Q_1@<<< 0@<<< 0 \\ @VVA\otimes dV @VVV @VVV\\ A\otimes Q_0@<<< 0@<<< 0 \end{CD}$$
Finally we have a morphism of bicomplexes $\sigma:P\otimes Q\to A\otimes Q$, defined by setting $\sigma_{0,n}:P_0\otimes Q_n\to A\otimes Q_n$ equal to $\epsilon\otimes Q_n$. The morphism of complexes $\operatorname {Tot}_\bullet^\oplus(\sigma):\operatorname {Tot}_\bullet^\oplus(P\otimes Q)\to\operatorname {Tot}_\bullet^\oplus(A\otimes Q)$ is defined by this property: $\operatorname {Tot}_n^\oplus(\sigma):\bigoplus_{i+j=n}P_i\otimes Q_j\to A\otimes Q_n$ is the map corresponding to $\epsilon\otimes Q_n$ on $P_0\otimes Q_n$ and to the zero map on the summands with $i\neq 0$. Our aim is to prove that $\operatorname {Tot}_\bullet^\oplus(\sigma)$ is a quasi-isomorphism, showing that its mapping cone $M_\bullet$ is exact.
Let's find $M_\bullet$ then. The module $M_n$ is $(A\otimes Q_n)\oplus(\bigoplus_{i+j=n-1}P_i\otimes Q_j)$, and the differential $M_{n+1}\to M_n$ is the map $$(A\otimes Q_{n+1})\oplus(\bigoplus_{i+j=n}P_i\otimes Q_j)\to (A\otimes Q_n)\oplus(\bigoplus_{i+j=n-1}P_i\otimes Q_j)$$ defined by the following data ($i_-$ is the canonical insertion of $-$):
- a map $A\otimes Q_{n+1}\to (A\otimes Q_n)\oplus(\bigoplus_{i+j=n-1}P_i\otimes Q_j)$ equal to $i_{A\otimes Q_n}\circ (A\otimes d)$.
- a map $P_0\otimes Q_n\to (A\otimes Q_n)\oplus(\bigoplus_{i+j=n-1}P_i\otimes Q_j)$ equal to $-i_{A\otimes Q_n}\circ (\epsilon\otimes Q_n)-i_{P_0\otimes Q_{n-1}}\circ (P_0\otimes d)$.
- for $i\neq 0$, a map $P_i\otimes Q_{n-i}\to (A\otimes Q_n)\oplus(\bigoplus_{i+j=n-1}P_i\otimes Q_j)$ equal to $-i_{P_{i-1}\otimes Q_{n-i}}\circ (-d\otimes Q_{n-i})+(-1)^{i+1}i_{P_i\otimes Q_{n-i-1}}\circ (P_i\otimes d)$.
So I'd say that $M_\bullet$ is $\operatorname{Tot}_\bullet^\oplus(B)$, where $B$ is the bicomplex below: $$\begin{CD} A\otimes Q_2 @<-\epsilon \otimes Q_2<< P_0\otimes Q_2@< -d\otimes Q_2<< P_1\otimes Q_2@<-d\otimes Q_2 <<P_2\otimes Q_2 \\ @VVA\otimes dV @VV-P_0\otimes dV @VVP_1\otimes dV @VV-P_2\otimes dV\\ A\otimes Q_1 @<-\epsilon \otimes Q_1<< P_0\otimes Q_1@< -d\otimes Q_1<< P_1\otimes Q_1@< -d\otimes Q_1<< P_2\otimes Q_1\\ @VVA\otimes dV @VV-P_0\otimes dV @VVP_1\otimes dV @VV-P_2\otimes dV\\ A\otimes Q_0 @<-\epsilon \otimes Q_0<<P_0\otimes Q_0@< -d\otimes Q_0<< P_1\otimes Q_0@< -d\otimes Q_0<< P_2\otimes Q_0 \end{CD}$$ $\operatorname{Tot}_\bullet^\oplus (B)$ is exact as all rows of $B$ are exact, since $-\otimes Q_n$ is an exact functor for any $n$. However it seems that Weibel, in the proof of Theorem 2.7.2, proposes to see $M_\bullet$ as $\operatorname {Tot}_\bullet^\oplus (B')[1]$, where the bicomplex $B'$ is obtained pasting $A\otimes Q_\bullet[-1]$, instead of $A\otimes Q_\bullet$, as follows: $$\begin{CD} A\otimes Q_1 @<d\circ (\epsilon \otimes Q_2)<< P_0\otimes Q_2@< d\otimes Q_2<< P_1\otimes Q_2@<d\otimes Q_2 <<P_2\otimes Q_2 \\ @VV-A\otimes dV @VVP_0\otimes dV @VV-P_1\otimes dV @VVP_2\otimes dV\\ A\otimes Q_0 @<d\circ(\epsilon \otimes Q_1)<< P_0\otimes Q_1@<d\otimes Q_1<< P_1\otimes Q_1@< d\otimes Q_1<< P_2\otimes Q_1\\ @VVV @VVP_0\otimes dV @VV-P_1\otimes dV @VVP_2\otimes dV\\ 0@<<<P_0\otimes Q_0@< d\otimes Q_0<< P_1\otimes Q_0@< d\otimes Q_0<< P_2\otimes Q_0 \end{CD}$$ This leaves me with several doubts, as in first place I'm not convinced that the rows of $B'$ are exact, and also I don't see how $\operatorname {Tot}_{n+1}^\oplus (B')$ can be viewed as the $M_n$ found above.
Another thing: a bit later, in Exercise 2.7.3, it is given some motivation on why we used $\operatorname {Tot}_\bullet^\oplus$ and not $\operatorname {Tot}_\bullet^{_\prod}$. But isn't it a matter of taste, since $B$ and $B'$ are bounded, hence $\operatorname {Tot}_\bullet^\oplus\cong \operatorname {Tot}_\bullet^{_\prod}$ canonically? Thanks in advance for any clarification.