Tensor Product of Complexes and the definition of the differentials Suppose we have the following complexes,
$$0 \rightarrow R \xrightarrow{x_1} R \rightarrow 0$$
$$0 \rightarrow R \xrightarrow{x_2} R \rightarrow 0$$
$$0 \rightarrow R \xrightarrow{x_3} R \rightarrow 0$$
where $R$ is a ring and $x_i \in R$ for all $i$. 
If $K.$ and $L.$ are two complexes, then the $n$th term in the tensor product is defined by $(K \otimes L)_n = \oplus_{p+q=n} K_p \otimes L_q$ and the maps are given by $\partial_n(a \otimes b) = \partial'(a) \otimes b + (-1)^p a \otimes \partial''(b)$, where $\partial'$, $\partial''$ are the differentials of the first and second complexes, respectively. So we get,
$$0 \rightarrow R \otimes R \otimes R \rightarrow (R \otimes R \otimes R) \oplus (R \otimes R \otimes R) \oplus (R \otimes R \otimes R) \rightarrow (R \otimes R \otimes R) \oplus (R \otimes R \otimes R) \oplus (R \otimes R \otimes R) \rightarrow R \otimes R \otimes R \rightarrow 0$$
which is obviously the same as 
$$0 \rightarrow R \xrightarrow{\partial_3} R \oplus R \oplus R \xrightarrow{\partial_2} R \oplus R \oplus R \xrightarrow{\partial_1} R \xrightarrow{\partial_0} 0$$
We have $\partial_3(1 \otimes 1 \otimes 1) = (x_1 \otimes 1 \otimes 1) + (1 \otimes x_2 \otimes 1) - (1 \otimes 1 \otimes x_3)$. We can of course write this down as $x_1 + x_2 - x_3$ since we are considering $R \otimes R \otimes R$ as just $R$.
Question
$x_1+x_2 - x_3$ is not contained in $R \oplus R \oplus R$, right? How is $\partial_3$ well-defined then? At first, I just thought of sending the coefficient of $x_1$ to the first summand, that of $x_2$ to the second summand, and so on.
But then I realized that $\partial_2$ is defined by sending $(a,b,c) \in R^3$ to $(\partial_2(a), \partial_2(b), \partial_2(c))$, right?
So how are these maps defined? 
Thanks in advance
 A: $\require{cancel}$Let me deal with the case of the tensor product of two complexes (I hope after that the case of three complexes should be easy). So let the first complex be $C = (0 \to R_1 \xrightarrow{x_1} R_0 \to 0)$, and the second one $C' = (0 \to R'_1 \xrightarrow{x_2} R'_0 \to 0)$; of course $R_0 = R_1 = R'_0 = R'_1 = R$, but this notation change will be helpful for what follows.
We want to compute $C \otimes C'$; it is equal to:
$$0 \to R_1 \otimes R'_1 \xrightarrow{d_2} (R_1 \otimes R'_0) \oplus (R_0 \otimes R'_1) \xrightarrow{d_1} R_0 \otimes R'_0 \to 0.$$
The differentials are computed as follows:
$$d_2(\underbrace{a \otimes b}_{\in R_1 \otimes R'_1}) = \underbrace{x_1 a \otimes b}_{\in R_0 \otimes R'_1} - \underbrace{a \otimes x_2 b}_{\in R_1 \otimes R'_0}$$
$$d_1(\underbrace{a \otimes b}_{\in R_1 \otimes R'_0}) = \underbrace{x_1 a \otimes b}_{\in R_0 \otimes R'_0} - a \otimes \cancel{d_0 b}$$
$$d_1(\underbrace{a \otimes b}_{\in R_0 \otimes R'_1}) = \underbrace{a \otimes x_2 b}_{\in R_0 \otimes R'_0} + \cancel{d_0 a} \otimes b$$
So now we can make all the identifications $R_0 = R_1 = R'_0 = R'_1 = R$ and use $R \otimes R = R$. We thus get a complex:
$$C \otimes C' = (0 \to R \xrightarrow{d_2} R^2 \xrightarrow{d_1} R \to 0),$$
where:
$$d_2(a) = (-x_2 a, x_1 a), \qquad d_1(a,b) = x_1 a + x_2 b.$$
