It seems that the definition of $T$ is unclear ("How is one supposed to interpret simplex $[w_0,w_0]$?") and that Hatcher's remark about its geometric motivation is somewhat obscure.
Indeed Hatcher's notation is sometimes a bit ambiguous and may therefore lead to confusion. Let me therefore begin with a few remarks which render more precisely what we need to understand Hatcher.
- What is a simplex in the sense of Hatcher?
For $n+1$ points $v_0,\ldots,v_n \in \mathbb R^m$ which do not lie in a hyperplane of $\mathbb R^m$ with dimension less than $n$, Hatcher defines $[v_0,\ldots,v_n] \subset \mathbb R^m$ as the convex hull of these points and calls it an $n$-simplex. Note, however, that Hatcher emphasizes
For purposes of homology it will be important to keep track of the order of the vertices of a simplex, so "$n$- simplex" will really mean "$n$-simplex with an ordering of its vertices".
In my opinion it is therefore more adequate to understand an $n$-simplex as an $(n+1)$-tuple $[v_0,\ldots,v_n]$ of points $v_i \in \mathbb R^m$ which do not lie in a hyperplane of $\mathbb R^m$ with dimension less than $n$. By an abuse of notation Hatcher also regards this tuple as the convex hull $[v_0,\ldots,v_n] \subset \mathbb R^m$ of these points (which is a topological space).
The standard $n$-simplex $\Delta^n$ is $[e_0,\ldots,e_n]$, where the $e_i$ are the standard basis vectors of $\mathbb R^{n+1} = \{(x_0,\ldots, x_n) \mid x_i \in \mathbb R \}$.
- What is $LC_n(Y)$?
For a convex set $Y$ in some Euclidean space, $LC_n(Y)$ denotes the subgroup of $C_n(Y)$ generated by the linear singular $n$-simplices $\sigma : \Delta^n \to Y$. A linear singular $n$-simplex $\sigma$ is uniquely determined by the images $w_i = \sigma(e_i)$ of the vertices $e_i$ of $\Delta^n$. In fact, linearity means that $\sigma(\sum_{i=0}^n t_ie_i) = \sum_{i=0}^n t_i\sigma(e_i) = \sum_{i=0}^n t_iw_i$. In other words, we have a bijective correspondence between linear $n$-singular simplices $\sigma : \Delta^n \to Y$ and $(n+1)$-tuples $[w_0,\ldots,w_n]$ of points $w_i \in Y$. The $w_i$ need not be distinct.
Via the above bijective correspondence we shall write linear singular $n$-simplices also in the form $[w_0,\ldots, w_n]$. This may a be source of confusion because$[w_0,\ldots, w_n]$ stands for both a tuple and a map. The map is given by
$$[w_0,\ldots, w_n](\sum_{i=0}^n t_ie_i) = \sum_{i=0}^n t_iw_i . \tag{1}$$
Expressions like $[w_0,w_0] \in CL_1(Y)$ make perfect sense. $[w_0,w_0]$ is the constant map $\Delta^1 \to Y$ with value $w_0$.
Of course each $n$-simplex $[v_0, \ldots, v_n]$ can be regarded as a linear singular $n$-simplex in a natural way via $(1)$. Let us denote by $LC^*_n(Y)$ the subgroup of $LC_n(Y)$ generated by the $n$-simplices in $Y$ (which simply means that all $v_i \in Y$). If $Y$ does not contain any $n$-simplex (which always happens for sufficiently large $n$), then $LC^*_n(Y) = 0$.
$LC_n(Y)$ and $LC^*_n(Y)$ are also defined for $n = -1$. In fact, these groups are $0$ because the set of generators is empty in both cases.
Note that the boundary operator $\partial$ maps $LC^*_{n+1}(Y)$ into $LC^*_n(Y)$ for all $n \ge -1$.
Based on his inductive definition of $T$, Hatcher proves the chain homotopy formula $\partial T + T \partial = \mathbb I - S$ by induction. This proof works without any geometric intuition.
However, Hatcher says
The geometric motivation for this formula is an inductively defined subdivision of $\Delta^n \times I$ obtained by joining all simplices in $\Delta^n \times \{0\} \cup (\partial \Delta^n) \times I$ to the barycenter of $\Delta^n \times \{1\}$, as indicated in the figure in the case $n = 2$.
What does this mean? Hatcher inductively constructs a subdivision of $\Delta^n \times I$ into a set of $(n+1)$-simplices. However, we need to do more. The simplices in the subdivision need to be endowed with a sign, i.e. we have to construct a certain chain $c(\Delta^n) \in LC^*_{n+1}(\Delta^n \times I) \subset LC^*_{n+1}(\mathbb R^{n+2})$.
Let us do it in a more general and formal way by considering $m$-simplices $[v_0, \ldots, v_m]$ in $\mathbb R^{n+1}$ and constructing a chain $c([v_0,\ldots,v_m]) \in LC^*_{m+1}\mathbb R^{n+2})$ such that the simplices occuring in this chain form a subdivision of the topological space $[v_0, \ldots, v_m] \times I$ and have coefficients $\pm 1$.
Technically we do this by induction on $m$ using barycentric subdivision to construct homomorphisms
$$c = c_m : LC^*_m(\mathbb R^{n+1}) \to LC^*_{m+1}(\mathbb R^{n+2}) .$$
Of course it suffices to define $c_m$ on the generators. For a point $x$ let us write $x^k = (x,k)$, $k =0,1$. The barycenter of $[v_0,\ldots,v_m]$ is denoted by $b_{[v_0,\ldots,v_m]} \in \mathbb R^{n+1}$.
$c_{-1}$ is the zero-map (which is the only homomorphism $0 \to LC^*_0(R^{n+2})$).
Given $c_m$, we define
$$c_{m+1}([v_0, \ldots, v_{m+1}]) = b_{[v_0, \ldots, v_{m+1}]}^1([v_0^0, \ldots, v_{m+1}^0] + c_m(\partial[v_0, \ldots, v_{m+1}])). $$
In this definition $b_{[v_0, \ldots, v_{m+1}]}^1$ is understood as the cone operator.
For $m = 0$ we get
$$c([v_0]) = [v_0^1,v_0^0]$$
because $\partial[v_0] = 0$ and $b_{[v_0]} = v_0$.
For $m =1$ we get
$$c([v_0,v_1]) = [b_{[v_0,v_1]}^1, v_0^0,v_1^0] + [b_{[v_0,v_1]}^1, v_1^0,v_1^1] - [b_{[v_0,v_1]}^1, v_0^0,v_0^1)] .$$
For $m = 2$ see Hatcher's figure. It is a chain of ten $3$-simplices.
All this applies to the standard $n$-simplex and gives a chain $c(\Delta^n) \in LC^*_{n+1}(\Delta^n \times I)$.
Hatchers continues:
What $T$ actually does is take the image of this subdivision under the projection $\Delta^n \times I \to \Delta^n$.
We can project the chain $c(\Delta^n)\in LC^*_{n+1}(\Delta^n \times I)$ onto $\Delta^n$ by replacing each vertex $x^k$ in a simplex of $c(\Delta^n)$ with $x$. This produces a chain $\bar c(\Delta^n) \in LC_{n+1}(\Delta_n)$. We have
$$\bar c([e_0]) = [e_0,e_0],$$
$$\bar c([e_0,e_1]) = [b_{[e_0,e_1]}, e_0,e_1] + [b_{[e_0,e_1]}, e_1,e_1] - [b_{[e_0,e_1]}, e_0,e_0] .$$
All linear singular simplices in $\bar c(\Delta^n)$ have images in $\Delta^n$, thus we can apply $\sigma : \Delta^n \to Y$ after each of these linear singular simplices and get the $(n+1)$-chain $\sigma \circ \bar c(\Delta^n) \in LC_{n+1}(Y)$.
This is what $T$ does: It associates to $\sigma \in LC_n(Y)$ the chain $T(\sigma) = \sigma \circ \bar c([e_0,\ldots,e_n]) \in LC_{n+1}(Y)$.
Whether this is a good "geometric motivation" is left to your judgement.
