Fundamental group of mapping cone of quotient map from suspension to reduced suspension
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 $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.