You are almost correct, the problem is in your assumption that a fine enough triangulation will yield that $G$ respects the boundary morphisms.
In fact we don't need to work cellularly, this works just as well with the singular chain complex (because $\Delta^n$ is contractible, so up to choosing a basepoint we have unique liftings $\Delta^n\to X$ to $\Delta^n\to X'$), and I'll rather use this to show how the differentials differ (it seems clearer to me that way)
Also, for simplicity, I'll take $\pi' = 0$ so $X' = \tilde X$.
Now we need to analyze the map $C_n(\tilde X)\otimes_{\mathbb Z[G]}\otimes M \to C_n(X;M)$. For each $\sigma : \Delta^n\to X$, fix one $\tilde \sigma : \Delta^n\to \tilde X$ that lifts it. By the usual lifting property, this is equivalent to choosing a lift $p_{\sigma}$ of $\sigma(e_0)$ for each $\sigma$, and this will come in handy.
This choice is what gives you the isomorphism $C_n(\tilde X) = \bigoplus_{\sigma : \Delta^n\to X}\mathbb Z[G]$
Now your map takes $\tau\otimes m$, writes it as $(g\cdot \tilde\sigma) \otimes m$ for some $g\in \pi, \sigma : \Delta^n \to X$, and then sends this to $\sigma \otimes g\cdot m$ (where I see $C_n(X;M) = C_n(X;\mathbb Z)\otimes M$). Note that $\sigma = p\circ \tau$ so it's almost $p\circ \sigma \otimes m$, but it's twister by the difference
Now look at $\partial (\tau\otimes m) = \partial \tau \otimes m = \sum_{i=0}^n (-1)^i d_i\tau \otimes m$, write $d_i\tau$ as $g_i\cdot (\widetilde{p\circ d_i\tau})$, and this is sent to $\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes g_i\cdot m$
Compare to what happens if you first send it over to $p\circ \tau\otimes g\cdot m$ ($g$ such that $g\cdot \widetilde{p\circ\tau} = \tau$), and then take the boundary : although clearly $d_i(p\circ \tau) = p\circ d_i\tau$, you have $g\cdot m$ instead of the varying $g_i$ : you get $\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes g\cdot m$
So a priori there's no reason for the two differentials to agree, they're somehow "twisted" by $\pi$, and that's precisely what local coefficients do : they twist the differential.
In fact, what you can do is, instead of fixing $p_\sigma$ for each $\sigma$, fix an antecedent $p_x$ for each $x\in X$. Then we define $\tilde\sigma$ to be the unique lift of $\sigma$ that sends $e_0$ to $p_{\sigma(e_0)}$ : this gives something more uniform, and you can check that in this situation, the differentials are almost the same : for $i\neq 0$, $g_i = g$, but for $i=0$, there's a problem in that $d_0\tau$ doesn't have the same evaluation in $e_0$ as $\tau$.
More precisely, let $\tau = \tilde\sigma$ (so that there's no need for a $g$). Then for each $i$, $d_i\tau$ is a lift of $d_i\sigma$, and for $i\neq 0$, $d_i\tau (e_0) = \tau(e_0) = p_{\sigma(e_0)}$, so that $d_i\tau = \widetilde{d_i\sigma}$ : there is no $g_i$. But for $i=0$, $d_0\tau(e_0) = \tau(e_1)$ which is not necessarily $p_{\sigma(e_1)}$ : it is related to it by some $g\in \pi$ (because it's also a lift of $\sigma(e_1)$), so that one of the differentials looks like $p\circ d_0\tau \otimes g\cdot m+ \sum_{i=1}^n (-1)^i p\circ d_i\tau \otimes m$, and the other is $\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes m$.
So they're almost the same except for a twist on the $0$th face - this is the same twist as the one in something called the standard resolution, when computing group (co)homology, if you know about that.
Now you can clearly (hopefully) see that there's no reasonable condition to impose on the triangulation or the cell-structure or whatever that will make sure that this twist disappears and that the two differentials actually agree.
Best Answer
Hatcher works with the cellular homology groups of CW-complexes. In Proposition 3B.1. he proves that the boundary map in the cellular chain complex $C_*(X×Y)$ is determined by the boundary maps in the cellular chain complexes $C_∗(X)$ and $C_∗(Y)$ via the formula $d(e_i×e_j) = de_i×e_j + (−1)^ie_i×de_j$.
Now consider $Y = S^k$ with the standard CW-structure (one $0$-cell $b^0$ and one $k$-cell $b^k$). Then $d = 0$ in $C_*(S^k)$. This is trivial for $k \ne 1$. For $k = 1$ we enconter a little problem because the cellular boundary formula (see p.140) requires to determine the degree of a map $S^0 \to S^0$. For the answer see In cellular homology, how should we define the degree of a map between $0$-spheres? It shows that $d=0$ also in this case.
Therefore $d(a \times b) = da \times b$ for the cells of $X \times S^k$.
The $n$-cells of $X \times S^k$ are the products $a \times b$ of all cells $a$ of $X$ and $b$ of $S^k$ such that $\dim a + \dim b = n$. Therefore the basis of the free abelian group $C_n(X \times S^k)$ is $$\mathcal B_n(X \times S^k) = \mathcal B_n(X) \times \{b^0\} \cup \mathcal B_{n-k}(X) \times \{b^k\} $$ where $\mathcal B_i(X)$ denotes the set of $i$-cells of $X$. Using the above boundary formula this shows that $$C_n(X \times S^k) \approx C_n(X) \oplus C_{n-k}(X).$$ Since