Totalizing a two-term twist action

The four summands and their degrees. Since $R_i$ has its two terms in degrees $-1$ and $0$ by [F2] and $X$ has its two terms in degrees $-1$ and $0$ by [F3], the index pairs $(p,q)$ with both $R_i^p$ and $X^q$ nonzero are $(-1,-1),(-1,0),(0,-1),(0,0)$, and the diagonal ${⨁}_{p+q=n}$ of [L1] collects them as $p+q=-2$ for $(-1,-1)$, as $p+q=-1$ for $(-1,0)$ and $(0,-1)$, and as $p+q=0$ for $(0,0)$; this gives the three displayed degrees with the four tensor summands $U_i{\otimes }_{A_m}{X}^{-1}$, $U_i{\otimes }_{A_m}{X}^0$, $A_m{\otimes }_{A_m}{X}^{-1}$, $A_m{\otimes }_{A_m}{X}^0$.

The differentials. By [L1] the differential on $U_i{\otimes }_{A_m}{X}^{-1}$ is $d_R\otimes 1+(-1{)}^{-1}1\otimes d_X={\beta }_i\otimes 1-1\otimes d$, which on an elementary tensor is $r\otimes x\mapsto {\beta }_i(r)\otimes x-r\otimes dx$, and the differential on $U_i{\otimes }_{A_m}{X}^0$ is ${\beta }_i\otimes 1+(-1{)}^{-1}1\otimes d_X={\beta }_i\otimes 1$, since $d_X=0$ on $X^0$ by [F3]; the differential on $A_m{\otimes }_{A_m}{X}^{-1}$ is $0+(-1{)}^{0}1\otimes d=1\otimes d$, that is $a\otimes x\mapsto a\otimes dx$, and on $A_m{\otimes }_{A_m}{X}^0$ it is $0$ because $R_i^1=0$ and $d_X^0=0$. No internal degree enters any sign, and all four maps preserve the total internal degree because ${\beta }_i$ and $d$ are degree-zero maps by [F2] and [F3].

The two routes cancel. For $r\otimes x\in U_i{\otimes }_{A_m}{X}^{-1}$ step 1.2 gives $d^{-2}(r\otimes x)={\beta }_i(r)\otimes x-r\otimes dx$, an element of the two summands of degree $-1$; applying $d^{-1}$ to the two pieces separately gives $d^{-1}({\beta }_i(r)\otimes x)={\beta }_i(r)\otimes dx$ in the first summand and $d^{-1}(-r\otimes dx)=-{\beta }_i(r)\otimes dx$ in the second, the latter because $d^{-1}$ on $U_i{\otimes }_{A_m}{X}^0$ is ${\beta }_i\otimes 1$ and $d_X(dx)=0$; the two results are negatives of one another, so $d^{-1}{d}^{-2}=0$ on the bottom term, and $d^0{d}^{-1}=0$ holds trivially because $d^0=0$. Hence the four displayed maps make the totalization a complex, as [L1] guarantees in general.

Unit form of the two upper summands. By [L4] the summands $A_m{\otimes }_{A_m}{X}^{-1}$ and $A_m{\otimes }_{A_m}{X}^0$ are degree-zero isomorphic to $X^{-1}$ and $X^0$ through the multiplication maps, so the middle term of the totalization may be written as $X^{-1}\oplus (U_i{\otimes }_{A_m}{X}^0)$ and the top term as $X^0$, with the differentials $x\mapsto dx$ and the $U_i$-component ${\beta }_i\otimes 1$ respectively.

Conclusion. A two-term twist complex $R_i$ and a two-term projective complex $X$ produce the totalization $U_i{\otimes }_{A_m}{X}^{-1}\to \left(U_i{\otimes }_{A_m}{X}^0\right)\oplus \left(A_m{\otimes }_{A_m}{X}^{-1}\right)\to A_m{\otimes }_{A_m}{X}^0$ with the four summands and the differentials of step 1.2, whose square vanishes by the explicit cancellation of step 2.1, and whose two $A_m$-columns may be read as $X^{-1}$ and $X^0$ by step 2.2. The sign $-1$ in the bottom differential is the Koszul sign $(-1{)}^p$ at $p=-1$, that is, it is attached to the homological degree of the first factor and not to any internal degree, which is the point of the construction.