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..
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 $\mathbb{Z}_2$.
Best Answer
"...if I subtract two elements of $\mathop{Ker} \partial_0$ I do not necessarily obtain $0$'' -- Correct, which is why the homology is not trivial: if $\sigma_1 - \sigma_2 \in \mathop{Im} \partial_1$ for all $\sigma_1$ and $\sigma_2$, then $[\sigma_1] = [\sigma_2]$ for all $\sigma_1$ and $\sigma_2$ so that $H_0$ would be $0$.