The concatenation $f_1 \cdot \ldots \cdot f_k$ is a map $I \to X$ obtained by dividing the interval into $k$ subintervals and doing $f_i$ on the $i$th subinterval. You can take these subintervals to be $[0, 1/k], [1/k, 2/k], \ldots, [(k-1)/k, 1]$, or you can take the product two loops at a time; for instance, $f_1 \cdot (f_2 \cdot (f_3 \cdot f_4)))$ would use the partition $[0, 1/2]$, $[1/2, 3/4]$, $[3/4, 7/8]$, $[7/8, 1]$. This is the partition giving the product $f_1 \cdot \ldots \cdot f_k$ that Hatcher refers to (and similarly for $f_1' \cdot \ldots \cdot f_l'$).
When Hatcher says that the $s$-partition on page 45 can be assumed to subdivide the partitions giving the products, he means that the partition points (e.g. the 0, 1/2, 3/4, 7/8, and 1 in my example) each appear as one of the values $s_0, s_1, \ldots s_m$. We can assume this because if our $s$-partition doesn't contain these points, we simply refine the $s$-partition by inserting them; any refinement of the $s$-partition will still have the desired property that the rectangles $R_{ij}$ each map into a single $A_{\alpha}$.
The generator $\beta$ is defined by the property that it is sent to a fixed generator of $H^{2n}(D^{2n},\partial D^{2n})$ by $H^{2n}(C_f,S^n) \to H^{2n}(D^{2n},\partial D^{2n})$, where the latter is induced by a map $(D^{2n},\partial D^{2n})\to (C_f, S^n)$ which is simply the attaching map.
But if $f\simeq g$, then you can construct a homotopy equivalence $C_f\to C_g$ which is compatible with these attaching maps, that is, such that the following square commutes up to homotopy
$$\require{AMScd}\begin{CD}(D^{2n},\partial D^{2n}) @>>> (C_f,S^n) \\
@VidVV @VVV\\
(D^{2n},\partial D^{2n}) @>>> (C_g,S^n)\end{CD}$$
that's essentially the content of proposition 0.18 (which Hatcher mentions in the paragraph you're referring to), or rather it follows from the proof of that proposition.
Since this square commutes up to homotopy, the corresponding square in cohomology commutes on the nose:
$$\begin{CD}H^{2n}(D^{2n},\partial D^{2n}) @<<< H^{2n}(C_f,S^n) \\
@AidAA @AAA\\
H^{2n}(D^{2n},\partial D^{2n}) @<<< H^{2n}(C_g,S^n)\end{CD}$$
which means that the generator for $C_g$ which is sent to the chosen generator of $H^{2n}(D^{2n},\partial D^{2n})$ is sent to the corresponding one for $C_f$, which is the claim.
Best Answer
The point is that we are considering these rings as graded commutative rings, and elements of even degree and elements of odd degree behave very differently in such rings. In particular, elements of even degree commute with all others, so they can generate a polynomial ring. On the other hand, two elements of odd degree anticommute, and in particular this means an element $\alpha$ of odd degree always satisfies $\alpha^2=-\alpha^2$. So, over a base ring in which $2\neq 0$, you cannot have a polynomial ring generated by an element in odd degree. In particular, then, this is why Hatcher doesn't write the cohomology ring of an odd-dimensional sphere as $\mathbb{Z}[\alpha]/(\alpha^2)$: there actually is no such thing as $\mathbb{Z}[\alpha]$ as a graded commutative ring, when $\alpha$ has odd degree.
As for why not to call the cohomology ring of an even dimensional sphere an exterior algebra, I'm guessing that Hatcher wants to be able to say that a tensor product of exterior algebras is an exterior algebra. This isn't true if you allow exterior algebras on generators of even degree, since the different generators will commute instead of anticommuting in the tensor product. Indeed, just as an element of odd degree cannot generate a polynomial ring (if $2\neq 0$), an element of even degree cannot be a generator of an exterior algebra unless there are no other generators (since it needs to anticommute with all the other generators, but it instead commutes with them since it has even degree).