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..
Pretty much everything you've said is correct. Because the boundary of an n-simplex is defined to be a sum of the restrictions of the map to various "faces" of the $n$-simplex, each of which is therefore an $n$-simplex...then if there's only one $0$-simplex, the boundary of the 1-simplex has to be $\sigma_0 \pm \sigma_0$. (And the sign is determined by the definition.)
Regarding $\sigma_2$ and orientation: No, it's not possible to orient it so that it gives a negative sign. The simplex $\sigma_2$ is a particular map from $\Delta^2$ (which I like to think of as $\{(x,y,z) \mid 0 \le x,y,z \le 1, x + y + z = 1 \}$) to the single point (call it $P$) defined by $(x, y, z) \mapsto P$. That's all it is. There's no "orientation" of the map.
It's possible to choose a different generator for $C_2$, namely $-\sigma_2$. But that doesn't change $\sigma_2$ itself.
Otherwise...you've got it exactly right.
Singular homology appears, at first, to be just weird. If you do simplicial homology, everything is nice and finite and manageable, so why go to the set of all possible continuous maps? Well you might ask. One answer is that after you prove a few more things, you'll find that the two techniques yield exactly the same results for simplicial complexes, so it doesn't matter. But that doesn't address "Why make it so complicated???" A good answer to that is "because not everything whose homology you wish to compute is actually as nice as a simplicial complex," so you either have to go through the (complicated, to me at least) simplicial approximation theorem, or you have to just say that this singular homology isn't looking so bad after all. And after you've done a few examples like this one, and the homology of a circle, and a few others, you rapidly learn how to do stuff where you never need to think about the singular simplices themselves, just as when you learn that an ordered pair is defined by
$$(a, b) := \{ \{a\}, \{a, b\}\},
$$
it seems really awkward and messy. But then you prove that $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$, and thereafter you use only that lemma, and never look at the formal set-theory definition again.
Best Answer
You confuse a lot of things. A singular $n$-simplex of $X$ is a continuous map $$\sigma_\alpha\colon \Delta^n \to X$$
The singular $n$-simplices of $X$ are the basis elements of the free abelian (chain) group $C_n(X)$, i.e. they generate $C_n(X)$.
A singular $0$-simplex is not being generated by singular $0$-simplices, but if your space $X$ admits precisely two singular $0$-simplices, then the chaingroup $C_0(X)$ is generated by these two $0$-simplices.
Also, it's certainly not true that you can somehow choose infinitely many generators for a chaingroup, say $C_1(X)$. The free abelian group $C_n(X)$ is a free $\mathbb{Z}$-module and its elements are formal sums $\sum_i n_i\sigma_i$ with $\mathbb{Z}$-coefficients $n_i$. These elements are called $n$-chains.
In summary, $n$-chains of $C_n(X)$ are linear combinations of basis elements the singular $n$-simplices $\sigma_\alpha$ that generate $C_n(X)$.
In order to get a better intuitive picture i would highly suggest you to look up the idea of homology, it's very well illustrated in Hatcher on p.99.