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 $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.
Your ideas are correct, but you have to make explicit how you regard an element of $\mathbb Z^{\mathbb N}$ as an element of $\pi_1(H)$ and that the resulting function $\phi : \mathbb Z^{\mathbb N} \to \pi_1(H)$ is injective. Let us elaborate 1.
Let us write $l_n^m : [0,1] \to K_n$ for the loop based at $0$ that travels counter-clockwise $m$ times around $K_n$. Explicitly, $l_n^m(t) = \frac{1}{n}(1- e^{2m\pi i t})$. Define
$$\psi((m_n)) : [0,1] \to H, \psi((m_n))(t) = \begin{cases}l_n^{m_n} (n(n+1)t - n) & t \in [\frac{1}{n+1},\frac{1}{n}] \\
0 & t = 0 \end{cases}$$
This is a well-defined continous map (since each neigborhood of $0$ contains all but finitely many $K_n$). Let $\phi((m_n)) = [\psi((m_n))]$, where $[-]$ denotes homotopy class of paths.
Let us show that $\phi$ is injective. There is a retraction $r_n : H \to K_n$ which maps all $K_r$, $r \ne n$, to $0$. Let $i_n : K_n \to H$ denote inclusion. The map $F_n = (r_n)_* \circ \phi$ has the property that the sequence $(m_n)$ is sent to the homotopy class of the path given by $l_n^{m_n} (n(n+1)t - n)$ for $t \in [\frac{1}{n+1},\frac{1}{n}]$ and maps all other $t$ to $0$. This path is clearly homotopic to $l_n^{m_n}$. Thus, if $\phi((m_n)) = \phi((m'_n))$, then $F_n((m_n)) = F_n((m'_n))$ for all $n$, i.e. $[l_n^{m_n}] = [l_n^{m'_n}]$ for all $n$. But this implies $m_n = m'_n$ for all $n$.
Identifying $\pi_1(K_n)$ with $\mathbb Z$ via the isomorphism $\iota_n : \mathbb Z \to \pi_1(K_n), \iota_n(m) = [l_n^m]$, we can express this alternatively as follows: The homomorphism
$$R : \pi_1(H) \to \mathbb Z^{\mathbb N}, R(u) = ((\iota_n)^{-1}(r_n)_*(u))$$
has the property $R \circ \phi = id$.
Also have a look at Fundamental group of mapping cone of quotient map from suspension to reduced suspension .
Best Answer
The same construction doesn't work for $SX$.
Note that the sequence of paths is typically combined into a single path by putting each path on a subinterval of $[0,1]$ with $1$ being a limit of those subintervals. Now with $\Sigma X$ we can map $1$ to the unique shared point. So every sequence covergent to $1$ will be mapped to a sequence converging to the unique point. But with $SX$ we can find a sequence converging to $1$ but its image converges to any point lying at the vertical line over $0$. There is no valid choice for value at $1$ making the construction continuous.
This also shows that any loop in $SX$ can go around only finitely many distinct subcircles of $SX$. You can conclude from that that the fundamental group is countable.
Let's dive into details. Let $SX=(X\times [0,1])/\sim$ and let $v_0=[(0,1)]_\sim$ be the top vertex. By $k$'th line I will understand the image of $\{1/k\}\times[0,1]$ in $SX$ and denote it by $L_k$. Note that $L_0$ will be the image of $\{0\}\times[0,1]$.
Your construction is as follows: for any sequence of naturals $n_1,n_2,\ldots$ let $f_k:[1/k,1/(k+1)]\to SX$ be a path such that $f(1/k)=f(1/k+1)=v_0$ and such that $f$ goes through $L_{n_k}$ line and back through say fixed $L_1$ line (so that they are pairwise non-homotopic). Finally we compose all $f_k$ into $f:[0,1]\to SX$ via $f(x)=f_k(x)$ if $x\in[1/k,1/(k+1)]$ and $f(0)=v_0$.
Note that this construction is continuous over $\Sigma X$ but not over $SX$. Indeed, let $w_k=[(1/k, 1/2)]_\sim$ and note that $w_k\to [(0,1/2)]_\sim\neq v_0$. But $f^{-1}(w_{n_i})$ is a single point that belongs to some $[1/t,1/(t+1)]$. So it forms a sequence convergent to $0$. This is a contradiction since the image does not converge to image of $0$ being $v_0$.
The main difference between $SX$ and $\Sigma X$ is that $\Sigma X$ is locally connected unlike $SX$. This implies that:
Proof. Assume that's not the case, so we have $L_{m_1},L_{m_2},\ldots$ fully contained in $im(f)$. Since $im(f)$ is compact then $$\overline{\bigcup_{i=1}^\infty L_{m_i}}\subseteq im(f)$$ This implies (by the intrinsic properties of $X$) that $L_0\subseteq im(f)$. But then $im(f)$ is not locally connected. Contradiction, since $f$ is a quotient map (onto its image) from a locally connected space (see this). $\Box$
Side note: another difference is that $SX$ is not an image of any path but $\Sigma X$ is (by the Hahn-Mazurkiewicz theorem, or by the mentioned construction).
Sketch of the Proof. There's a countable number of subcircles (basically a subcircle is a pair $(L_i,L_j)$ of lines) in $SX$. Since every path goes around only finitely many of them then it means that to any path we can associate a sequence $(n_1,n_2,\ldots)\in\mathbb{Z}^\infty$ of corresponding winding numbers. Only finitely many entries are non-zero. And there are only countable many such sequences. $\Box$