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 $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.
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.
Best Answer
For the sake of more transparent formulae let us use the equivalent definition $$\Sigma X = X \times J /S$$ with $J =[-1,1]$ and $S = \{*\} \times J \cup X \times \{-1,1\}$.
Let $\partial J^2 = \{-1,1\} \times J \cup J \times \{-1,1\}$ denote the boundary of $J^2$. Then
$$\Sigma^2 X = X \times J^2 / R $$ with $R = \{*\} \times J^2 \cup X \times \partial J^2$.
For any boundary preserving homotopy $H : J^2 \times I \to J^2$ (which means that $H(\partial J^2 \times I) \subset \partial J^2$) define $$H^* = id_X \times H : X \times J^2 \times I \to X \times J^2 .$$ We have $H^*(R \times I) \subset R$, thus $H^*$ induces a homotopy $$\tilde H : \Sigma^2 X \times I \to \Sigma^2 X .$$
It therefore suffices to show that the switch map $$\tau : J^2 \to J^2, \tau(s,t) = (t,s)$$ and the reflection map $$\rho : J^2 \to J^2, \rho(s,t) = (s,-t)$$ are homotopic via a boundary preserving homotopy. Note that $\tau, \rho$ are boundary preserving (which means that they map $\partial J^2$ into itself).
We can do this by proving that $\phi = \tau \circ \rho^{-1}$ and the identity on $J^2$ are homotopic via a boundary preserving homotopy. Note that $$\phi(s,t) = (-t,s) .$$
Geometrically this a counterclockwise $90^0$-rotation around the center $0$ of the square $J^2$. I shall not write down an explicit homotopy. This is an easy exercise based on the standard homeomorphism
$$h : J^2 \to D^2 .$$ See for example Homeomorphism between the unit disc and the unit square.