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.