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.
You seem to claim that $SX$ is homotopy equivalent to the second space in your picture (which I shall denote by $S'X \subset \mathbb R^2$). This is not true. The yellow circle does not belong to $S'X$, thus $S'X$ is not compact. If you have any map $f : SX \to S'X$, then its image is compact and therefore must be contained in some $S'_n = \bigcup_{i=1}^n A_i$. This is a finite wedge of circles. We have $f = i_n f_n$ where $f_n : SX \to S'_n$ is the restriction of $f$ and $i_n : S'_n \to S'X$ denotes inclusion. If $g : S'X \to SX$ would be a homotopy inverse of $f$, then the identity on $\pi_1(SX)$ would factor through $\pi_1(S'_n)$ which is false.
However, there is no reason to replace $SX$ by another space. By the way, note that $\Sigma X$ is known as the Hawaiian earring. In Hatcher's Example 1.25 it is denoted as "The Shrinking Wedge of Circles".
As basepoint for $SX$ choose the midpoint $x_0$ of the black line segment and as basepoint for $\Sigma X$ choose the cluster point $y_0$ of the circles $B_i$. We have obvious pointed retractions $r_i : SX \to A_i$ (which project $A_j$ to the black line segment for $j \ne i$) and $s_i : \Sigma X \to B_i$ (which map $B_j$ to $y_0$ for $j \ne i$). This gives us group homomorphisms
$$\phi : \pi_1(SX,x_0) \to \prod_{i=1}^\infty \pi_1(A_i,x_0) = \prod_{i=1}^\infty \mathbb Z, \phi(a) = ((r_1)_*(a), (r_2)_*(a),\ldots),$$
$$\psi : \pi_1(\Sigma X,y_0) \to \prod_{i=1}^\infty \pi_1(B_i,y_0) = \prod_{i=1}^\infty \mathbb Z, \psi(b) = ((s_1)_*(b), (s_2)_*(b),\ldots) .$$
It is easy to see that $\psi$ is surjective, but $\phi$ is not. In fact, $\text{im}(\phi) = \bigoplus_{i=1}^\infty \mathbb Z$. This is true because all but finitely many $(r_i)_*(a)$ must be zero (otherwise the path representing $a$ would run infinitely many times through both endpoints of the black line segment, thus would have infinite length).
Obviously we have $\psi \circ q_* = \phi$, where $q : SX \to \Sigma X$ is the quotient map.
Now let us apply van Kampen's theorem. Write $C = U_1 \cup U_2$, where $U_1$ is obtained from $C$ by removing the tip of mapping cone and $U_2$ by removing the base $\Sigma X$. Both $U_k$ are open in $C$. We have
$U_1 \cap U_2 \approx SX \times (0,1) \simeq SX$ (thus $U_1 \cap U_2$ is path connected)
$U_1 \simeq \Sigma X$ (in fact, $\Sigma X$ is a strong defornation retract of $U_1$)
$U_2$ is contractible.
We conclude that $\Phi : \pi_1(U_1) * \pi_1(U_2) = \pi_1(\Sigma X) * 0 = \pi_1(\Sigma X) \to \pi_1(C)$ is surjective. Its kernel $N$ is the normal subgroup generated by the words of the form $(i_1)_*(c)(i_2)_*^{-1}(c)$, where $i_k : U_1 \cap U_2 \to U_k$ denotes inclusion and $c \in \pi_1(U_1 \cap U_2)$. Since $(i_2)_*^{-1}(c) = 0$, we see that $N$ is the normal closure of the image of the map $(i_1)_* : \pi_1(U_1 \cap U_2) \to \pi_1(U_1)$. But under the identifications $U_1 \cap U_2 \simeq SX$ and $U_1 \simeq \Sigma X$ we see that $(i_1)_*$ corresponds to $q_* : \pi_1(SX) \to \pi_1 (\Sigma X)$.
Hence $\pi_1(C) \approx \pi_1 (\Sigma X)/ N'$, where $N'$ is the normal closure of $\text{im}(q_*)$.
The surjective homomorphism $\psi' : \pi_1(\Sigma X) \stackrel{\psi}{\rightarrow} \prod_{i=1}^\infty \mathbb Z \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$ has the property $\psi' \circ q_* = 0$, thus $\text{im}(q_*) \subset \ker(\psi')$. Since $\ker(\psi')$ is a normal subgroup, we have $N' \subset \ker(\psi')$, thus $\psi'$ induces a surjective homomorphism $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.
Best Answer
Instead of $S^n$ and $SS^n$ we consider $S^{n-1}$ and $SS^{n-1} \approx S^n$.
On $D^n \times \{-1,1\}$ define $(x,-1) \sim (x,1)$ for $x \in S^{n-1}$. Then $\Sigma^n = (D^n \times \{-1,1\})/\sim \phantom{.}$ is the quotient space obtained by gluing two copies of $D^n$ along their boundary. Let $p : D^n \times \{-1,1\} \to \Sigma^n$ denote the quotient map. Clearly $\Sigma^n \approx S^n \approx SS^{n-1}$. This gives us an identification $$h : \Sigma^n \stackrel{\approx}{\to} SS^{n-1} .$$ Let $\xi_0 = (0,\ldots,0,1) \in S^{n-1}$ and $L$ be the line segment connecting $0$ and $\xi_0$. The images of $(0,\pm 1)$ under $h$ are the two "suspension points" of $SS^{n-1}$ and the image of $L' = p(L \times \{-1,1\})$ under $h$ is the line segment $\{\xi_0\} \times I \subset SS^{n-1}$.
The quotient map $q : D^n \to D^n/L$ maps $S^{n-1}$ homeomorphically onto $q(S^{n-1})$ and by an abuse of notation we write $S^{n-1} = q(S^{n-1}) \subset D^n/L$. Thus we can identify $SS^{n-1}/\{\xi_0\} \times I$ with the quotient space obtained from $(D^n/L) \times \{-1,1\}$ by identifying $(x,-1)$ and $(x,1)$ for $x \in S^{n-1}$.
We shall show that there exists a homeomorphism $h : D^n/L \to D^n$ such that $h([x]) = x$ for $x \in S^{n-1}$. This gives us a homeomorphism $(D^n/L) \times \{-1,1\} \to D^n \times \{-1,1\}$ which is compatible with the quotient maps and thus gives us the desired homeomorphism $SS^{n-1}/\{\xi_0\} \times I \to SS^{n-1}$.
Write the points $x \in D^n$ as $x = (\xi,t)$ with $\xi \in D^{n-1}$ and $t \in [-1,1]$ where $\lVert \xi \rVert^2 + t^2 \le 1$. Define $d(\xi) = \sqrt{1 - \lVert \xi \rVert^2}$ and $$\phi : D^n \to D^n, \phi(\xi,t) = \begin{cases} (\xi,(1 - \lVert \xi \rVert)d(\xi) + t\lVert \xi \rVert) & t \ge 0 \\ (\xi,(1 - \lVert \xi \rVert)d(\xi) + t(2-\lVert \xi \rVert)) & t \le 0 \end{cases}$$ Note that both parts of the definition produce the same value for $t = 0$. Moreover we have in fact $\phi(\xi,t) \in D^n$ since each line segment $S(\xi) = \{\xi\} \times [-d(\xi),d(\xi)] = D^n \cap (\{\xi\} \times [-1,1])$ is mapped by $\phi$ onto itself: The line segment $S^+(\xi) = \{\xi\} \times \left[0,d(\xi)\right]$ is mapped by $\phi$ linearly onto the line segment $\{\xi\} \times \left[(1- \lVert \xi \rVert) d(\xi),d(\xi)\right]$ and the line segment $S^-(\xi) = \{\xi\} \times \left[-d(\xi),0\right]$ is mapped by $\phi$ linearly onto the line segment $\{\xi\} \times \left[-d(\xi),(1- \lVert \xi \rVert) d(\xi)\right]$; this means $\phi(S(\xi)) = S(\xi)$. Note that $\phi$ is bijective on $S(\xi)$ if $\xi \ne 0$ because then $0 \le (1- \lVert \xi \rVert) d(\xi) < 1$. For $\xi = 0$ the map $\phi$ stretches $S^-(0)$ to $S(0)$ and collapses $L = S^+(0)$ to $\xi_0$.
Thus $\phi$ induces a map $h : D^n/L \to D^n$ which is a homeomorphism as desired.
We have considered a special $\xi_0 \in S^{n-1}$. Given an arbitrary $x_0 \in S^{n-1}$ we can find an orthogonal linear map $\phi : \mathbb R^n \to \mathbb R^n$ such that $\phi(x_0) = \xi_0$. It induces a homeomorphism $\phi' : S^{n-1} \to S^{n-1}$ which reduces the general case of an arbitrary $x_0 \in S^{n-1}$ to the above special case.