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..
I have not read Hatcher, but Prism's go back to Eilenberg-Steenrod. The idea is simple if $H(x,t)$ is a homotopy on $\Delta_p$ (just assume for the moment it is constant on the boundary), then we can view $H$ as a function on the prism $I\times \Delta_p$. Now the boundary of the prism can be thought of as the difference of the top and the bottom, (the sides are constant). This can be then extended to sums of simplices. And so it shows that homotopy gives a chain-homotopy which induces and isomorphism.
I think the problem arises because $I\times \Delta_p$ is no longer a simplex.
Athough personally I have never really understood why this is such a big issue since one can easlily and explicitly write a prism as a sum of simplices.
Best Answer
Short answer
This triangulation is a particular case of a general construction for simplicial sets, and triangulations of products of simplices $\Delta^p\times \Delta^q$ are related to combinatorial objects named shuffles. A reference for that is probably Gabriel, Zisman, "Calculus of Fractions and Homotopy Theory", chapter II, or the modern monograph "Simplicial Homotopy Theory" by Goerss and Jardine (though I am not sure they discuss shuffles there).
A longer explanation
Let me try to explain from scratch how these triangulations arise. My explanation is not for the connoisseurs of simplicial stuff, so I apologize for a lengthy post and sweeping under the rug some details.
Let's say that a triangulation is something consisting of the following data:
We also require by definition each face to be triangulated, i.e. if $\sigma\in F$ and $\tau \subset \sigma$ is a nonempty subset of $\sigma$, then $\tau \in F$.
Now for each $n = 0,1,2,\ldots$ let $K_n$ be the set of the $n$-simplices of the triangulation, where by an $n$-simplex we mean an ordered collection of $n+1$ vertices $v_0 \to \cdots \to v_n$: $$K_n = \{ (v_0 \to \cdots \to v_n) \mid \{ v_0,\ldots,v_n \} \in F \}.$$ Here we do not require $v_i$ to be different! They are ordered, but neighboring vertices may repeat. Treating something spanned by $(n+1)$ vertices with repetitions as an $n$-simplex may seem odd, but it will be very useful in a moment.
We have special maps $\sigma_i\colon K_n\to K_{n+1}$, called degeneracy operators. By definition, $\sigma_i$ repeats the $i$-th vertex: $$\sigma_i (v_0 \to \cdots \to v_n) = (v_0 \to \cdots \to v_{i-1} \to v_i \to v_i \to v_{i+1} \to \cdots \to v_n).$$ These operators, together with face operators $\partial_i\colon K_n\to K_{n-1}$ (removing the $i$th vertex) satisfy the so-called simplicial identities, and the sets $K_n$ with these operators form a simplicial set (you can read the details elsewhere), which I will denote by $K$.
When some face lies in the image of a degeneracy operator $\sigma_i$ (i.e. has repeating vertices), we say that it is degenerate.
For example, a triangle with ordered vertices $0 < 1 < 2$ has a triangulation where $F$ consists of all nonempty subsets of $\{ 0, 1, 2 \}$.
Here I enumerated the simplices in dimensions $0,1,2$ and I highlighted the degenerate ones (with repetitions of vertices). Note that starting from dimension $3$, everything will be degenerate (since we have only three vertices).
Now comes the interesting part: if $K$ and $K'$ are two simplicial sets, I can take their product $K\times K'$, where the $n$-simplices in $K\times K'$ will be pairs $$(K\times K')_n = K_n\times K_n' = \{ (x,x') \mid x\in K_n, ~ x'\in K_n' \},$$ and the degeneracy operators will be $$\sigma_i (x,x') = (\sigma_i (x), \sigma_i (x')).$$
Now a simplex $(x,x')$ is degenerate in $K\times K'$ if it is in the image of $(\sigma_i, \sigma_j)$ for some $i=j$, i.e. if both $x$ and $x'$ are degenerate, and via the same degeneracy operator $\sigma_i$. So $K\times K'$ will have more nondegenerate simplices than $K$ and $K'$. Degenerate things in $K$ and $K'$ may give something nondegenerate in $K\times K'$, and intuitively, these are the extra parts we have to add to triangulate $K\times K'$.
Let's work out the easiest example: take the interval $\Delta^1$ which consists of two ordered vertices $0\to 1$. The product $\Delta^1 \times \Delta^1$ will have the following simplices:
Again, I highlight the degenerate simplices. We have four $0$-simplices, five (!) nondegenerate $1$-simplices, and two (!) nondegenerate $2$-simplices. For instance, $(0\to 0\to 0, 0\to 0\to 1)$ is degenerate because it is $\sigma_0 (0\to 0,0\to 1)$. However, $(0\to 0\to 1, 0\to 1\to 1)$ is nondegenerate: though both $0\to 0\to 1$ and $0\to 1\to 1$ are degenerate in $\Delta^1$, they are degenerate via different operators $\sigma_i$ and $\sigma_j$ with $i\ne j$.
From this data I can draw a picture with triangulation, and only nondegenerate simplices will matter. I get a square with vertices $(0,0), (0,1), (1,0), (1,1)$, precisely the $0$-simplices of $\Delta^1\times \Delta^1$. Then the nondegenerate $1$-simplex $(0\to 0, 0\to 1)$ becomes the interval connecting $(0,0)$ with $(0,1)$; the nondegenerate $1$-simplex $(0\to 1, 0\to 1)$ becomes the "diagonal" connecting $(0,0)$ with $(1,1)$, and so on. The nondegenerate $2$-simplex $(0\to 0\to 1, 0\to 1\to 1)$ becomes the triangle spanned by the vertices $(0, 0)$, $(0, 1)$, $(1, 1)$, and $(0\to 1\to 1, 0\to 0\to 1)$ becomes the triangle spanned by the vertices $(0, 0)$, $(1, 0)$, $(1, 1)$.
Similarly one could work out $\Delta^2\times \Delta^1$ (this example looks more exciting, but going through all the possible simplices is kind of tedious):
Though (hopefully) one can understand from the example above how to construct triangulations from a list of nondegenerate simplices in $\Delta^p\times \Delta^q$, I haven't really explained how an abstract simplicial set gives rise to some CW-complex (glued precisely from the nondegenerate simplices). This is called geometric realization, but that's a different story...
And for more details on combinatorics of $\Delta^p\times \Delta^q$, I refer to Gabriel and Zisman. They explain how the nondegenerate simplices in $\Delta^p\times \Delta^q$ precisely come from shuffles.
Triangulations of prisms and chain homotopies
You mentioned that to prove that homotopic maps $X\to Y$ induce chain homotopic maps between complexes $C_\bullet (X) \to C_\bullet (Y)$, you need to triangulate $\Delta^n\times \Delta^1$. In fact, you can construct a chain homotopy inductively, avoiding pulling out of a hat a mysterious combinatorial expression for $h_n$ for all $n$. See Tammo tom Dieck, "Algebraic Topology", §9.3.