On the meaning of formal sums of $k$-cubes, i.e. $k$-chains (in integration on manifolds) A singular $k$-cube in $A \subseteq \mathbb R^n$ is a continuous function $c : [0,1]^k \to A$. A singular $0$-cube in $A$ is then a function $f : \{0\}\to A$, what amounts to the same thing, a point in $A$. A formal sum of singular $k$-cubes in $A$ multiplied by integers is an $k$-chain, i.e. sums like
$$
 3c_1 + 2c_2 - 4c_3
$$
for singular $k$-cubes $c_1, c_2, c_3$. These could be defined more formally as "polynomials" in the $k$-cubes on $A$, i.e. functions from the $k$-cubes to $\mathbb Z$ for which only finitely many function values are non-zero.
By the way these definitions are taken from Spivak, Calculus on Manifolds, page 97 (the more formal one is from an exercise therein).
So now my question. As formal sums they do not admit any interpretation behind being merely formal objects with prescribed rules to manipulate them. Now for an $n$-cube (and also an $n$-chain by linear extension) its boundary is defined as such a formal sum of $n-1$-cubes got from this cube by fixing one variable. 
Then relations like $\partial^2 = 0$, or extending the integral over $n$-chains lineary from $n$-cubes, all work out fine, and ultimately lead to Stokes theorem. But when I look at the definition, and try to give it some meaning, or picture to have, such a formal sum seems to correspond to an union of sets (i.e. the parts of the boundary) enriched by some notion of orientation (by the sign in the sum) and "speed of transversal" (by its magnitude) (Remark: Actually this seems more of a multiset, as different $n$-cubes could share parts as sets).
So, it might be possible to interpret these sums pointwise, i.e. for example $2c_1 - c_2$ not as a formal sum, but as the pointwise addition of functions, i.e. $2\cdot c_1(x) - c_2(x)$ for all $x \in [0,1]^k$ if $c_1, c_2$ are singular $k$-cubes. Of course, I mean this addition if we interpret every element of $\mathbb R^n$ as a vector (i.e. a position vector w.r.t. the origin), as a priori points could not be added.
Comparing with the picture (I appended one from the above mentioned book) this seems to work fine. So why not use this more concrete definition, but use abstract formal sums? (or is my interpretation to naive, and fails somewhere?)

 A: Let's see what happens when we interpret these sums pointwise. For example, $\partial I^2$ is an 1-chain given by
$$\partial I^2 = -I^2_{(1,0)} + I^2_{(1,1)} + I^2_{(2,0)} - I^2_{2,1}$$
Now evaluating this "sum" at a point $x \in [0,1]$ gives
$$\partial I^2 = -(0,x) + (1, x) + (x, 0) - (x, 1) = (1, -1) $$
Try the same calculation for $\partial I^3$ and you'll get $(1,-1,1)$. Clearly this doesn't make sense. 
To see how Spivak "summed" the $(n-2)$-chain in his proof of $\partial^2=0$, let's take $\partial^2I^2$ as an example. Now we must sum the boundaries of each $I^2_{(i, \alpha)}$ with the correct orientation.
So, for instance,
\begin{align}
\partial (-I^2_{(1,0)})(0) 
&= -[-(I^2_{(1,0)})_{(1,0)}(0) + (I^2_{(1,0)})_{(1,1)}(0)] \\
&= -[-I^2_{(1,0)}(I^1_{(1,0)}(0)) + I^2_{(1,0)}(I^1_{(1,1)}(0))] \\
&= -[-I^2_{(1,0)}(0) + I^2_{(1,0)}(1)] \\
&= -[-(0,0) + (0,1)]\\
&= (0,0) + - (0,1)
\end{align}
We don't add these: they are just "oriented" points in $\mathbb{R}^2$.
Similarly, we get
\begin{align}
\partial(I^2_{(1,1)})(0) &= -(1,0) + (1,1)\\
\partial(I^2_{(2,0)})(0) &= -(0,0) + (1,0)\\
\partial(-I^2_{(2,1)})(0) &= (0,1) + -(1,1)
\end{align}
Each of the four corners of the square are appear twice with opposite signs.
For instance,
$(0,1) = (I^2_{(1,0)})_{(1,1)} = (I^2_{(2,1)})_{(1,0)}$ (this is the statement $(I^n_{(i,\alpha)})_{(j,\beta)} = (I^n_{(j+1,\beta)})_{(i,\alpha)}$ in Spivak's proof). But it's orientation is negative in the former and positive in the latter. So they cancel each other out - they are not "added".
To summarize, we don't add two different n-cubes in an n-chain. We add the coefficients of the same n-cube if it appears more than once.
