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For example,

UPDATE: One way to write this is in degree a group homology style. Then we can view elements of $1$ \Gamma(C)_n$ as decorated with these $n$-tuple paths, the width and height determine which group we have land in, and the boundary map takes an alternating sum of deleting commas with the understanding that $(C_1 NN = EE = 0$. So, for example, taking the boundary of this element in $C_3 \otimes D_0) D_2$:$$d(a \oplus otimes b)_{(N,E,N,NE)} = (C_1 da) \otimes D_1) b_{(E,N,NE)} - a \oplus (C_0 otimes b_{(NE,N,NE)} + a \otimes D_1)$, corresponding to N, NE, and E respectively. consider the tuple $(N,E)$ in degree $2$, which comes from b_{(N,NE,NE)} - a factor $C_1 \otimes D_1$b_{(N,E,NNE)}.The face map $d_0$ induces $d \otimes id$ to the "E" factor; the face map (The last term involving $d_1$ induces the projection NNE$ is dropped because it is zero. The first term had a boundary map applied to the "NE" first factor ; the face map because it was $d_2$ induces zero.N$.)
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The face maps have two characters. The map $d_i$ for $i > 0$ simply deletes the element $i$ from the ordered set $[n]$, and reindexes; if the resulting map $[n-1] \to [k]$ is no longer surjective, the corresponding factor maps to zero. By contrast, the map $d_0$ deletes $0$ and reindexes, but if the corresponding map $\phi$ is no longer surjective its image is isomorphic to $[k-1]$, and we apply the boundary map.

Unfortunately, a chain complex isn't very useful without its differential, and that's more complicated to describe. The boundary map is the alternating sum of face maps; each face map deletes $i$ from the ordered set $0 < \cdots < n$ and reindexes. If $i > 0$ and one of the resulting projections to $[p]$ or $[q]$ is no longer surjective, the corresponding factor maps to zero; if $i = 0$ then one or both of the maps to $[p]$ or $[q]$ misses zero, the appropriate image(s) are isomorphic to $[p-1]$ or $[q-1]$, and we apply the boundary map on those factors.

For example, in degree $1$ we have $(C_1 \otimes D_0) \oplus (C_1 \otimes D_1) \oplus (C_0 \otimes D_1)$, corresponding to N, NE, and E respectively. consider the tuple $(N,E)$ in degree $2$, which comes from a factor $C_1 \otimes D_1$. The face map $d_0$ induces $d \otimes id$ to the "E" factor; the face map $d_1$ induces the projection to the "NE" factor; the face map $d_2$ induces zero.
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The bad news is that in degree $n$, this tensor product has $3^n$ terms.

The functor $\Gamma$ can be roughly described as follows. If we write $[n]$ for the ordered set $0 < 1 < \cdots < n$, then $$ \Gamma(C)_n = \bigoplus_{k} \bigoplus_{\phi\colon [n] \twoheadrightarrow [k]} C_k. $$ When you take the tensor product of $\Gamma(C)$ and $\Gamma(D)$ levelwise, you get a direct sum indexed by pairs of surjections $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$.

The functor $N$ then will take the quotient of this by the subcomplex of degenerate ones; those where the maps $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$ factors through a surjection $[n] \twoheadrightarrow [m]$. In practice, the pairs which are not degenerate are those for which the map $[n] \to [p] \times [q]$ is injective.

As a result, we have that $$ N(\Gamma(C) \otimes \Gamma(D))_n = \bigoplus_\phi C_p \otimes D_q $$ where the sum is indexed by injections $[n] \to [p] \times [q]$ where composing with either projection is surjective.

In practice, you can index this direct sum by $n$-tuples of strings of elements from${N, NE, E}$, representing a path of length $n$; $p$ and $q$ are determined by the height and width of the path.

Unfortunately, a chain complex isn't very useful without its differential, and that's more complicated to describe.