Thank you to Mariano Suárez-Alvarez. I think the cohomology ring will work, somthing like this:
We know that the cohomology ring of a genus $g$ surface has the property that any non-zero class in degree one has a class it can cup with to get a non-trivial element of $H^2$ (this could be from direct computation or by Poincare duality.) In the wedge sum, we notice that WLOG $X$ must have trivial $H^2$. Thus any two degree 1 class supported on $X$ cup to 0, and any class supported on $X$ cups to zero with a class supported on $Y$. Thus $X$ has trival $H^1$.
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
$\bigvee_{2g}S^2\to \Sigma X_g\to S^3$ is a cofibration sequence.
Moreover, $H_2(\Sigma X_g) = H_1(X_g) = \mathbb Z^{2g}$ and $H_3(\Sigma X_g)= H_2( X_g) = \mathbb Z$, and the higher $H_n$'s are zero
$\Sigma X_g$ is simply connected, and so is the space you want to compare it to, so it suffices to find a map which induces an isomorphism in homology.
It's clear that the map $\bigvee_{2g}S^2\to \Sigma X_g$ already induces an isomorphism on $H_2$, so it suffices to find a map $S^3\to \Sigma X_g$ which induces an isomorphism on $H_3$. This map will not come from a map $S^2\to X_g$ by suspension, because $X_g$ has no higher $\pi_i$'s (if I remember correctly, orientable surfaces are always $K(\pi,1)$'s)
So for this, we may want to use the better version of Hurewicz, which not only says that $\pi_2\to H_2$ is an isomorphism (because $\Sigma X_g$ is simply-connected), but also that $\pi_3\to H_3$ is an epimorphism (see here, at "absolute version")
This in particular means that there is $S^3\to \Sigma X_g$ such that $H_3(S^3)\to H_3(\Sigma X_g)$ sends $1$ to $1$, in particular it's an isomorphism on $H_3$, and we get what we wanted