This claim occurs in the proof of Proposition 4F.1 in the version from 2002 (section "Spectra and Homology Theories").
It is not true, because in general $\Sigma^i X \vee \Sigma^i Y$ has cells of dimension $ > 2i-1$.
However, Hatcher only wants to prove that
$(\ast)$ $\pi_{n+i}(\Sigma^i X \vee \Sigma^i Y) \approx \pi_{n+i}(\Sigma^i X \times \Sigma^i Y)$ for $n+i < 2i-1$.
But now Tyrone's comment applies.
By the way, the following statement is correct and suffices to prove $(\ast)$:
$\Sigma^i X \vee \Sigma^i Y$ and $\Sigma^i X \times \Sigma^i Y$ have the same $(2i-1)$-skeleton.
The cells of $ \Sigma^i X \times \Sigma^i Y$ have the form $e \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. We have $\dim(e \times e') \le 2i-1$ only in case $e = \{ \ast \}$ and $\dim(e') \le 2i-1$ or $e' = \{ \ast \}$ and $\dim(e) \le 2i-1$ because $\Sigma^i Z$ has one $0$-cell $\{ \ast \}$ (the basepoint) and all other cells have dimension $\ge i$.
$\Sigma^i X \vee \Sigma^i Y = \Sigma^i X \times \{ \ast \} \cup \{ \ast \} \times \Sigma^i Y \subset \Sigma^i X \times \Sigma^i Y$ consists of all cells $e \times \{ \ast \}, \{ \ast \} \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. They have dimension $\le 2i-1$ if and only if their nontrivial factor $e$ resp. $e'$ has dimension $\le 2i-1$.
This proves the statement. Note that it can be generalized to the following:
If $X, Y$ are CW-complexes with one $0$-cell (being the base point) and no cells of dimension $1,\dots,i-1$, then $X \vee Y$ and $X \times Y$ have the same $(2i-1)$-skeleton.
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:
Lemma. Let $f:[0,1]\to SX$ be a continuous function. Then there are at most finitely many $k$ such that $L_k\subseteq im(f)$.
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).
Conclusion: $\pi_1(SX)$ is countable.
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. $\Box$
Best Answer
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 $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.