Cellular Homology via Stable Homotopy one can define cellular homology by letting
$C_n(X)=\{\mathbb{S}^n, X^n/X^{n-1}\}$, where $X$ is a CW complex and the curly brackets mean stable homotopy classes of maps. Now the differential of the resulting complex is supposed to be given by a map $X^n/X^{n-1}\to\Sigma X^{n-1}/X^{n-2}$ and I struggle to understand this map.
The only candidate I can think of would be the suspension of an attaching map. More precisely, on an $n$-cell, we use the inverse of the characteristic map to end up in $\mathbb{S}^n$, identify this with $\Sigma\mathbb{S}^{n-1}$ via a (canonical) homeomorphism, then apply the attaching map to end up in $\Sigma X^{n-1}$. 
If this is correct (is it?), I still do not see why this is indeed a differential, i.e. d^2=0. 
Furthermore, I would like to do concrete calculations with this formulation, if necessary only in very low dimensions (say, 1 or 2). So in particular I would like to know how the ordinary cellular boundary operator can be recovered from the fancy one above. Does anyone know a detailed reference for this view on cellular homology or can provide any useful insights?
Thank you.
 A: I am not completely clear on the motivation of your question, but I think I can answer the question as-asked.  In my opinion, this is easiest to see with a complicated diagram.  Start with the inclusions of subcomplexes:
$$\begin{array}{ccccccccccc}X^1 & \to & X^2 & \to & \cdots & \to & X^{n-1} & \to & X^n & \to & X^{n+1} & \to & \cdots\end{array},$$
and then throw in each quotient map dangling off:
$$\begin{array}{ccccccccccc}X^1 & \to &  X^2 & \to & \cdots & \to & X^{n-1} & \to & X^n & \to & X^{n+1} & \to & \cdots \\ \downarrow p_1 & & \downarrow & & & & \downarrow p_{n-1} & & \downarrow p_n & & \downarrow p_{n+1} & &\\ X^1 & & X^2 / X^1 & & & & X^{n-1} / X^{n-2} & & X^n / X^{n-1} & & X^{n+1} / X^n & & \cdots,\end{array}$$
where I've elected to name the downward maps for later use.
Each of these is what's called a cofiber sequence, where a cofiber sequence is a pair of sequential maps $A \to B \to C$ with the property that a test map $T \to B$ can be lifted to a commuting triangle $T \to A \to B$ if and only if the composite $T \to B \to C$ is null-homotopic.  The way you get cofiber sequences in homotopy theory is by attaching cones to subspaces ($A \to B \to B \cup_A Cone(A)$), which for nice enough subspaces agrees with the quotient sequence $A \to B \to B/A$.  An important lemma is that the maps in cofiber sequences satisfy the differential property:

Setting $T = A$, the map $A = T \to B$ lifts to a map $A = T \to A$ by taking the identity, and this forces (by the "only if") the long composite $A \to B \to C$ to be null.

An exceedingly useful fact about cofiber sequences is that they can be extended: in the sequence of maps $$\begin{array}{ccc} A & \to & B \\ & \to & B \cup_A Cone(A) \\ & \to & (B \cup_A Cone(A)) \cup_B Cone(B) \simeq \Sigma A \\ & \to & \Sigma A \cup_{B \cup_A Cone(A)} Cone(B \cup_A Cone(A)) \simeq \Sigma B \\ & \to & \cdots,\end{array}$$
any adjacent pair forms a cofiber sequence.  Abbreviating all these messy cones, one says that $A \to B \to C$ extends to $$A \to B \to C \to \Sigma A \to \Sigma B \to \Sigma C \to \Sigma^2 A \to \cdots.$$
Now what does this have to do with you?  Well, we have a whole bunch of cofiber sequences in that diagram with all the quotients of subcomplexes, and if we extend them, we find long sequences of the form $$X^{n-1} \to X^n \xrightarrow{p_n} X^n / X^{n-1} \xrightarrow{\partial_n} \Sigma X^{n-1} \to \Sigma X^n \to \cdots,$$ where I've again decided to give a name for later use to the "new map" that comes from extension.  You'll notice that the subcomplexes $X^n$ themselves sit in a remarkable position in the diagram: they have a map to the right to $X^{n+1}$ and a map down to $X^n / X^{n-1}$, both of which sit in different cofiber sequences.  Now, here's an important definitional assertion:

Your boundary map $X^{n+1} / X^n \to \Sigma(X^n / X^{n-1})$ is the same as the composite $(\Sigma p_n) \circ \partial_{n+1}$.  That is: start at your favorite node on the bottom row, follow the map $\partial_{n+1}$ to move diagonally back up to (a suspension of) the top row, then follow $p_n$ down to move back to the bottom row.

Now that we've said all these things about cofiber sequences, the assertion that this gives a differential follows quickly: performing this pair of operations twice yields a four-fold composite $((\Sigma^2 p_{n-1}) \circ (\Sigma \partial_n)) \circ ((\Sigma p_n) \circ \partial_{n+1})$.  Since we can associate composition, you find $\Sigma \partial_n$ and $\Sigma p_n$ right next to each other in the middle --- and these are two maps which appear adjacent to each other in the cofiber sequence $$X^{n-1} \to X^n \xrightarrow{p_n} X^n / X^{n-1} \xrightarrow{\partial_n} \Sigma X^{n-1} \to \cdots.$$  Hence, by the boxed lemma, their composite is zero, and hence the four-fold composite must also be zero, and that is exactly the differential condition on your boundary operator.  Since the map of spaces $X^{n+1} / X^n \to \Sigma^2 X^{n-1} / X^{n-2}$ is itself null, applying any decent functor (such as homotopy groups) will also yield zero.
You can easily dress this up to fit into, e.g., the equivariant setting --- the important thing was just that cofiber sequences were around to work with.  Moreover, if you know anything about spectral sequences, this is identical to the argument that the $d^1$-differential of the associated filtration spectral sequence is in fact a differential.  Let me know if I've missed your point, and I'll revise the answer accordingly.
