Hint for the first question. Given the cone $\Gamma X$ of a topological space $X$, you can cut it in the middle : $\Gamma X \simeq X_1 \cup X_2$ with $X_1 \simeq \Gamma X$, $X_2 \simeq X\times[0,1]$ and $X_1 \cap X_2 \simeq X$.
Given $f,g \colon X \to Y$ homotopic by $H \colon X \times [0,1] \to Y$, you could now be able to define a map $$M_g \to M_f$$ which will be (easily) a homotopy equivalence.
First consider the homology of the pair $(X^i \cup A^{i+1}, X^{i-1} \cup A^i)$. This is a CW pair, so by excision $$H_i(X^i \cup A^{i+1}, X^{i-1} \cup A^i) \cong \tilde{H}_i(X^i \cup A^{i+1} / X^{i-1} \cup A^i) \cong \tilde{H}_i (X^i / X^{i-1} \cup A^i) \cong H_i(X^i, X^{i-1} \cup A^i)$$where the second isomorphism holds because adding the $(i+1)$-cells of $A$ will have no effect on $H_i$ since we're collapsing the $i$-skeleton of $A$ to a point in the quotient (this is basically a relative version of Lemma 2.34 in your book).
This means that we can substitute this into the long exact sequence of the pair $(X^i \cup A^{i+1}, X^{i-1} \cup A^i)$ to get
$$\cdots \to H_i(X^{i-1} \cup A^i) \to H_i(X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1} \cup A^i) \to H_{i-1}(X^{i-1} \cup A^i) \to \cdots.$$
But $X^i \cup A^{i+1}$ fits into another long exact sequence, too: the long exact sequence of the pair $(X^{i+1} \cup A^{i+2}, X^i \cup A^{i+1})$, which is given by
$$\cdots \to H_{i+1}(X^{i+1} \cup A^{i+2}) \to H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i \cup A^{i+1}) \to \cdots$$
where we've identified the relative group in the middle by the same isomorphism above with $i$ replaced by $i+1$.
so define the map $H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1}\cup A^i)$ to be the composition $H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1} \cup A^i)$, where the first map is from the second long exact sequence above, and the second map is from the first long exact sequence above.
The remaining steps are to show that the sequence of these maps forms a chain complex and that the homology of this chain complex yields the homology of the pair $(X,A)$.
Best Answer
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.