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..
Let me use $(v_0,\ldots,v_n)$ to denote the unoriented $n$-simplex given by the vertices $v_0,\ldots,v_n$, and $[v_0,\ldots,v_n]$ the oriented $n$-simplex with the orientation determined by the given ordering.
For this triangulation, there exists a way to orient the $2$-simplices in the complex so that the resulting edge orientations do not contradict each other. Simply take the $2$-simplex spanned by $(v_0,v_1,v_2)$ and orient it counterclockwise; I will denote this as $[v_0,v_2,v_1]$, but any cyclic permutation of the vertices gives the same oriented $2$-simplex. (In general cyclic permutations might not result in the same orientation. It works here because $3$-cycles are even permutations.)
Then the edge spanned by $(v_0,v_2)$ has the orientation $[v_0,v_2]$ (going tail to head; note that a cyclic permutation for this oriented $1$-simplex doesn't give the same orientation, because $2$-cycles are odd permutations). Now orient the $2$-simplex spanned by $(v_0,v_2,v_3)$ so that the induced orientation on $(v_0,v_2)$ is $[v_0,v_2]$. Namely, if we give $(v_0,v_2,v_3)$ the clockwise orientation $[v_0,v_2,v_3]$ then this induces the orientation $[v_0,v_2]$ on $(v_0,v_2)$ as desired.
Best Answer
It might be easier to see why we would use alternating sums via the map $\partial_1:C_1\to C_0$, i.e. the map sending lines to vertices, and then it should be clear that the parity of the vertex is not as important as the fact that the boundary map must be expressed in terms of an alternating sum. Suppose we are looking at a triangle (or a 2-simplex), with vertices $v_0,v_1,v_2$. Then, $C_1=\mathbb{Z}\Big\{\{v_0,v_1\},\{v_1,v_2\},\{v_2,v_0\}\Big\}$, where the line $\{v_i,v_j\}$ represents the line that begins at $v_i$ and ends at $v_j$ . Now, the boundary map formula tells us that $$\partial_1(\{v_0,v_1\})+\partial_1(\{v_1,v_2\})+\partial_1(\{v_2,v_0\})=v_1-v_0+v_2-v_1+v_0-v_2=0 $$ The fact that the sum is zero tells us that these lines form the boundary of a shape that is a dimension higher (i.e. the face of the triangle). In fact, any element $[f]\in C_n$ such that $\partial_n[f]=0$ is called a boundary. The only way that the sum could equal zero was because it was alternating. If we defined the boundary map by instead negating when we removed even terms, then the above sum would still be $0$. This idea will still hold in higher dimensions, i.e, using parity of a vertex is a very natural way to define an alternating sum - that is why it is important.