I don't know how explicit we can go, but I'll give it a try. We have to go first through the homotopy-theoretical part.
Since $\{ * \} \subseteq X, \{ * \} \subseteq Y$ are cofibrations, $X \vee Y \subseteq X \times Y$ also is. Let $Z$ be a pointed space and consider the long exact sequence of homotopy for the pair $X \vee Y \subseteq X \times Y$, ie. the sequence
$\ldots \rightarrow [\Sigma ^{2}(X \vee Y), Z] \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X \vee Y), Z] \rightarrow [X \wedge Y, Z] \rightarrow \ldots$,
where $[-,-]$ is the pointed set of homotopy classes of basepoint-preserving maps. Note that for any $n \geq 0$, $\Sigma^{n}(X \vee Y)$ is homeomorphic to $\Sigma^{n}X \vee \Sigma ^{n} Y$. I will not distinguish between the two.
Let $k \geq 1$ and define a map
$\psi ^{k}: \Sigma^{k}(X \times Y) \rightarrow \Sigma^{k}X \vee \Sigma^{k}Y$
$\psi ^{k} = \Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})$,
where $\pi: X \times Y \rightarrow X, Y$ are the projections and $i: X, Y \rightarrow X \vee Y$ are the inclusions. Addition is performed via the suspension structure on $\Sigma^{k}(X \times Y)$, so this is why we require $k \geq 1$. (Observe that even though I denote it by addition this is not necessarily commutative for $k=1$.)
If $j: X \vee Y \hookrightarrow X \times Y$ is the inclusion, then I claim that $\psi ^{k}$ is the left inverse to $\Sigma^{k}j$, ie. $\psi ^{k} \circ \Sigma^{k}j = id_{\Sigma^{k}(X \vee Y)}$. This is important because $\Sigma^{k}j$ are connecting maps in the long exact sequence of homotopy. Indeed, one computes
$\psi ^{k} \circ (\Sigma^{k}j) = (\Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})) \circ \Sigma^{k}j = \Sigma^{k}(i_{X} \pi_{X} j) + \Sigma^{k}(i_{Y} \pi _{Y} j) = \Sigma^{k}(id_{X} \vee const) + \Sigma^{k}(const \vee id_{Y}) \simeq (\Sigma^{k}id_{X} + const) \vee (const + \Sigma^{k}id_{Y}) \simeq \Sigma^{k}id_{X} \vee \Sigma^{k}id_{Y} \simeq id_{\Sigma^{k}X \vee \Sigma^{k}Y}$.
(One can also see this geometrically.) This immediately implies that for all $k \geq 1$ and all $Z$ the $[\Sigma^{k}(X \times Y), Z] \rightarrow [\Sigma^{k}(X \vee Y), Z]$ induced by $j$ is surjective and - by exactness of the long exact sequence - that for all $n \geq 1$ the map $[\Sigma^{n}(X \smash Y), Z] \rightarrow [\Sigma^{n}(X \times Y), Z]$ has zero kernel. In particular, for $k=1$ we have the short exact sequence of groups
$0 \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow 0$
Moreover, the map induced by $\psi^{1}$ splits it and shows that there is a natural isomorphism
$\phi: [\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow [\Sigma(X \times Y), Z]$,
of groups, where the product is only semi-direct, because our groups are not necessarily abelian. This is enough for our purposes, since we also have natural bijections
$[\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y), Z] \times [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), Z]$.
(The second one follows from from the fact that $\vee$ is the direct sum in the category of pointed spaces.) Yoneda lemma establishes that there is an isomorphism
$\theta: \Sigma(X \times Y) \rightarrow _{\simeq} \Sigma(X \smash Y) \vee \Sigma(X) \vee \Sigma(Y) $
in the homotopy category of pointed spaces, ie. a homotopy equivalence that we were after. It takes a little bookkeeping in the above Yoneda-lemma argumentation to see that such map is given by
$\theta = \Sigma(p) + \psi^{i} = \Sigma(p) + \Sigma^{1}(i_{X} \pi_{X}) + \Sigma^{1}(i_{Y} \pi_{Y})$,
where $p: X \times Y \rightarrow X \wedge Y$ is the natural projection. (This is what we get if we start with $id \in [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ and trace it back by all the bijections above to $[\Sigma(X \times Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ - and this is the way to discover the isomorphisms "hidden" by Yoneda lemma.)
I understand that my exposition is far from perfect, but if you would like me to go into more detail over some parts, please comment.
To answer first your questions. A) The statement is correct, and B) your proof is correct, including step $3)$.
I'm not sure of a good reference (I'd guess Hatcher has a thing to say about it, though), but a way to see what the attaching map is is as follows.
We know that $S^1\times S^1$ is a CW complex consisting of two 1-cells and a single 2-cell. Thus we have a cofibration sequence
$$S^1\xrightarrow{w}S^1\vee S^1\rightarrow S^1\times S^1$$
where $w$ is the attaching map and represents a class in $\pi_1(S^1\vee S^1)$. Using the Seifert-Van Kampen theorem we find that $\pi_1(S^1\vee S^1)\cong \mathbb{Z}\ast\mathbb{Z}$ is the free group on the two generators, $a=in_1:S^1\hookrightarrow S^1\vee S^1$, $b=in_2:S^1\hookrightarrow S^1\vee S^1$ given by the inclusions of the two respective wedge summands. Thus the homotopy class $w=a^{i_1}b^{j_1}\dots a^{i_r}b^{j_r}$ is some word in the generators $a,b$.
Now the abelianisation of this is just the free abelian group on these two generators $(\pi_1(S^1\vee S^1))_{ab}\cong\mathbb{Z}\oplus\mathbb{Z}$, and the Hurewicz theorem tells you that this is isomorphic to the homology $H_1(S^1\vee S^1)$. Now $H_1(S^1\times S^1)\cong\mathbb{Z}\oplus\mathbb{Z}$ also, and the inclusion of the 1-skeleton $S^1\vee S^1\hookrightarrow S^1\times S^1$ induces an isomorphism on homology. We see this using the homology exact sequence of the cofibration above
$$0\rightarrow H_2(S^1\times S^1)\xrightarrow{\cong} H_1(S^1)\xrightarrow{w_*} H_1(S^1\vee S^1)\xrightarrow{\cong} H_1(S^1\times S^1)\rightarrow 0.$$
The right hand map must be an isomorphism since both involved groups are free abelian and there is no torsion in the sequence. It follows that the left-hand map is also an isomorphism, since $H_2(S^1\times S^1)\cong\mathbb{Z}$ and $H_1(S^1)\cong\mathbb{Z}$ (they are oriented manifolds 2- and 1-manifolds respectively). Thus $w_*=0$ on homology.
The point of introducing the Hurewicz theorem earlier was that we may use its naturality to get a commutative diagram
$\require{AMScd}$
\begin{CD}
@>>>0@>>>\pi_1S^1@>\pi(w)>>\pi_1(S^1\vee S^1)@>>>\pi_1(S^1\times S^1)@>>>0\\
@V V V @VV V@VV \cong V @V V(-)_{ab}V@VV\cong V\\
0@>>> H_2(S^1\times S^1)@>>>H_1S^1@>w_*>>H_1(S^1\vee S^1)@>\cong>> H_1(S^1\times S^1)@>>>0
\end{CD}
Here the vertical maps are all Hurewicz morphisms. All three are abelianisations, and two turn out to be isomorphisms, using that $\pi_1(S^1\times S^1)=\pi_1S^1\oplus\pi_1S^1\cong\mathbb{Z}\oplus\mathbb{Z}$. $\pi(w)$ is the homomorphism induced by $w$. The bottom row of this diagram is exact, although the top is not necessarily so.
Since $\pi_1S^1$ is free abelian the homomorphism $\pi(w)$ will be determined by what it does on the generator, which is the identity $id_{S^1}$. Note that
$$\pi(w)[id_{S^1}]=[w\circ id_{S^1}]=[w].$$
Now we observed above that $w_*=0$, so the commutative diagram above tells us that $(\pi(w))_{ab}=0$. Thus $w$ lies in the kernel of the abelianisation homomorphism $ab:\mathbb{Z}\ast\mathbb{Z}\rightarrow \mathbb{Z}\oplus\mathbb{Z}$, which sends the free generators $a,b$ to the free abelian generators of the same name. The kernel of this map is the subgroup generated by commutators $[x,y]=xyx^{-1}y^{-1}$ in the words.
The point is that while $w$ is non-trivial in the non-abelian group $\pi_1(S^1\vee S^1)$, after suspending it becomes trivial in the abelian group $\pi_2(\Sigma (S^1\vee S^1))$ (recall that $\pi_2$ is always abelian).
In fact if $X$ is a path connected space then the Hurewicz map $h_1:\pi_1(X)\rightarrow H_1X$ is an abelianisation. However, after suspension $\pi_1\Sigma X=0$ so the Hurewicz map $h_2:\pi_2X\xrightarrow{\cong} H_1X$ is now an isomorphism. The suspension homomorphism $\Sigma:\pi_1X\rightarrow\pi_2\Sigma X$ actually identifies with the abelianisation, and this follows from naturality $h_2\circ \Sigma=\sigma\circ h_1$, where $h_2$ is the isomorphism above, $\sigma:H_1X\xrightarrow{\cong} H_2\Sigma X$ is the homology suspension, which is also an isomorphism, and $h_1$ is abelianisation.
Hence $\Sigma w=0\in\pi_2(\Sigma(S^1\vee S^1))=\pi_2(S^2\vee S^2)\cong\mathbb{Z}\oplus\mathbb{Z}$.
Incidentally, the map $w$ is a so-called Whitehead product, and takes the form
$$w=aba^{-1}b^{-1}.$$
You can see it has this form explicitly by identifying the torus $S^1\times S^1$ as a quotient of $I\times I$. The identification traces around the oriented boundaries, as indicated in Hatcher pg. 5.
More explicitly, Hatcher describes the product cell structure on a product of CW complexes on pg. 8. Then the cell structure on $S^1$ with one $0$-cell and one $1$-cell is given by identifying the two points in the boundary $S^0\cong \partial I$. Choosing a relative homeomorphism $(D^2,S^1)\cong (I\times I,I\times\partial I\cup \partial I\times I)$ and keeping track of the orientations it is clear that $w$ indeed has the form described.
The Whitehead product has higher dimensional generalisations that give attaching maps for the top cells of all products $S^n\times S^m$, and indeed more generally for any product of the form $\Sigma X\times \Sigma Y$.
Best Answer
Notation: let $\bar\alpha:D^{k+l}\to S^{k}\times S^l$ denote the characteristic map associated with attaching the top cell in $S^k\times S^l$. Let $i_{S^k}, i_{S^l}:S^k,S^l\to S^k\vee S^l$ and $j:S^k\vee S^l\to S^k\times S^l$ be standard inclusions, and let $\pi_{S^k},\pi_{S^l}:S^k\times S^l\to S^k, S^l$ be projections.
Using the notation of the first answer in the post you linked to, we know that the map $$\psi:=\Sigma(i_{S^k}\pi_{S^k})+\Sigma(i_{S^l}\pi_{S^l}):\Sigma(S^k\times S^l)\to \Sigma(S^k\vee S^l)$$ satisfies $$\psi\circ \Sigma j\simeq \text{id}_{\Sigma(S^k\vee S^l)}$$ by cogroup structures of reduced suspensions.
The relationship between the attaching map and the characteristic map suggests that $$\Sigma\bar\alpha\circ \Sigma i=\Sigma j\circ\Sigma\alpha$$ where $i:S^{k+l-1}\to D^{k+l}$ is the inclusion of boundary. Composing with $\psi$, we get $$\ast\simeq \ast\circ\Sigma i\simeq(\psi\circ\Sigma\bar\alpha)\circ \Sigma i\simeq (\psi\circ\Sigma j)\circ \Sigma \alpha\simeq \Sigma\alpha$$ since $\psi\circ\Sigma\bar\alpha$ is a map from $D^{k+l}$ which is contractible. This shows that $\Sigma\alpha$ is null-homotopic. (and of course the top cell of $S^k\times S^l$ splits off after suspension)