Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn't be an answer...)
First some preliminary notions:
For a topological space $X$, an $n$-simplex in $X$ is a continuous map $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.
An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th vertex of $n_+$.
Then for the two homologies:
The singular (unreduced) chain complex on a space $X$, is the chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0$$
where $C_n(X)$ is the free abelian group $\mathbb{Z}[S_n(X)]$ generated by the set $S_n(X) = \{ \sigma : \Delta^n \to X \}$ of all $n$-simplices in $X$ (i.e. the set of all continuous maps $\Delta^n \to X$). The boundary maps $\partial_n : C_n(X) \to C_{n-1}(X)$ is given by $\partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X$.
The $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th singular homology group of $X$.
A simplicial complex $S$ is a set $S = \bigcup_{n=0}^{\infty} S_n$ where $S_n = S(n_+)$ being a set of ordered $n$-simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. The simplicial chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0$$
consists of the free abelian groups $C_n(S) = \mathbb{Z}[S_n]$ generated by the $n$-simplices. The boundary map $\partial_n : C_n(S) \to C_{n-1}(S)$ is given by $\partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma$ where $d_i = S(d^i) : S_n \to S_{n-1}$ is the face maps $d_i(\sigma) = \sigma \circ d^i$.
The $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th simplicial homology group of $S$.
Lastly we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$, where $d_i \sigma \times \Delta^{n-1}$ is identified with the $i$'th face of $\sigma \times \Delta^n$.
Then if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex).
NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..
Although you do not mention it in your question, it it obvious that you are interested in the simplicial homology of $\Delta$-complexes.
As an $n$-simplex Hatcher understands an ordered $n$-simplex which is an $(n+1)$-tuple $[v_0,\ldots,v_n]$ of vertices $v_i$. This means that an $n$-simplex contains more information than the set $\{v_0,\ldots,v_n\}$ of its vertices - in fact, if we take different orderings of the set of vertices, then this yields different $n$-simplices. Do not confuse this with the concept of an oriented simplex which is usually defined as an equivalence class of ordered simplices, two ordered simplices being equivalent if they originate from each other by an even permutation of their vertrices (i.e. we have $[v_0,\ldots,v_n] \sim [v_{\pi(0)},\ldots,v_{\pi(n)}]$ for each even permutation $\pi$).
The boundary homomorphism $\partial_n : \Delta_n(X) \to \Delta_{n-1}(X)$ is defined on the generators $\sigma^n : [v_0,\ldots,v_n] \to X$ by
$$\partial_n(\sigma^n) = \sum_{i=0}^n (-1)^n \sigma^n \mid [v_0,\ldots,\hat{v}_i,\ldots,v_n] .$$
The ordered $(n-1)$-simplices $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ are the faces of $[v_0,\ldots,v_n]$. More precisely, $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$ is the $i$-th face of $[v_0,\ldots,v_n]$. In the above formula it is essential that we associate the sign $(-1)^i$ to the $i$-th face $[v_0,\ldots,\hat{v}_i,\ldots,v_n]$. Only these signs allow to show that $\partial_{n-1}\partial_n = 0$.
If you work with unordered $n$-simplices, i.e. with the sets $\{v_0,\ldots,v_n\}$, then we obtain of course a set of $n+1$ unordered $(n-1)$-simplices $\{v_0,\ldots,\hat{v}_i,\ldots,v_n\}$ which we may call the faces of $\{v_0,\ldots,v_n\}$, but we do not have any chance to reasonably define the notion of an $i$-th face of the set $\{v_0,\ldots,v_n\}$.
The geometric meaning of orderings of vertices if explained on p. 105.
Best Answer
By definition, $C_k(X)$ is a vector space whose basis is the set of $k$-simplices of $X$. When $X=D^0$, $X$ has a single $0$-simplex and no $k$-simplices for $k\neq 0$. This means that $C_0(D^0)$ is one-dimensional (i.e., isomorphic to $\mathbb{Z}_2$) and $C_k(D^0)$ is trivial if $k\neq 0$.
Now we want to compute $H_k(D^0)$. That means we want to look at the maps $\partial_k:C_k(D^0)\to C_{k-1}(D^0)$ and $\partial_{k+1}:C_{k+1}(D^0)\to C_k(D^0)$ and compute the kernel of $\partial_k$ mod the image of $\partial_{k+1}$. Now the definitions of these maps are pretty complicated, but fortunately we don't have to worry about the definitions in this case because our vector spaces are so trivial. If $k>0$, then $C_k(D^0)$ is trivial. Since $\ker\partial_k$ is a subspace of $C_k(D^0)$, $\ker\partial_k$ is also trivial, and since $H_k(D^0)$ is a quotient of $\ker\partial_k$, $H_k(D^0)$ is also trivial. This proves $H_k(D^0)=0$ for $k\neq 0$.
Now let's consider $k=0$. The map $\partial_0:C_0(D^0)\to C_{-1}(D^0)$ has domain $\mathbb{Z}_2$ and codomain $0$ (we usually don't talk about $C_{-1}(D^0)$; it is always just $0$ because there is no such thing as a $(-1)$-simplex). So $\partial_0$ must send everything to $0$, so its kernel is all of $\mathbb{Z}_2$. On the other hand, the map $\partial_1:C_1(D^0)\to C_0(D^0)$ must be $0$, since its domain is $0$. So the image of $\partial_1$ is trivial. Thus $H_0(D^0)$ is $\mathbb{Z}_2$ mod the trivial subspace, or just $\mathbb{Z}_2$.