Fix $n, m\in\mathbb{N}$. We use $D_n$ to denote the disjoint union of two $(n-1)$-dimensional closed disks. For each $1\leq i \leq m$, we use $S(n, i)$ to denote $S^n$. Then clearly $D_n$ can be embedding into $S(n, 1)$ and we put $O_1 = S(n,1)$. If $m>1$, we then can let $S(n, 2)$ intersect $O_1$ so that the intersection is equal to $D_n$. We use $O_2$ to denote the union of $S(n, 1)$ and $S(n, 2)$ intersecting this way. We can further let $S(n, 3)$ intersect $O_2$ such that again the intersection is equal to (the embedding of) $D_n$. In this way we have $O_m$ for all $m\in\mathbb{N}$.
As an example, when $n=1$, $D_1$ only consists of two disjoint points and $O_2$ looks like the following:
When $n=1$, $O_m$ is $m$ copies of circles glued together at the top and the bottom point. In this case, $O_m$ is a one-dimensional CW complex that consists of two disjoint points and $2m$ many lines. Since any one of the lines is a contractible subcomplex, $O_m$ is homotopic equivalent (via the quotient mapping) to the usual wedge sum of $2m-1$ copies of circles. According to wiki (or this post), when $n=1$, the fundamental group of $O_m$ is the free group with $2m-1$ generators. In general, when $n>1$, what is the fundamental group of $O_m$?
If your construction is equivalent to the one mentioned in freakish's comment, then your space $O(n, m + 1)$ is obtained from $O(n, m)$ by taking the disjoint union of $O(n, m)$ and a copy of $S^n$, and then identifying a copy of $D^{n - 1} \sqcup D^{n - 1}$ embedded in each. I claim that $O(n, m + 1) \simeq O(n, m) \vee S^n \vee S^1$, so that $O(n, m) \simeq S^n \bigvee_{\text{$(m - 1)$ copies}} (S^1 \vee S^n)$, and so the fundamental group is, as mentioned on the Wikipedia page you linked, the free product of the fundamental groups of the wedge summands. That is, for $n = 1$, the free group on $2m - 1$ generators, and for $n > 1$, the free group on $m$ generators.
The homotopy equivalence is not difficult to see, but would be irritating to write down explicitly. For the case $n = m = 2$, it looks something like this.
Below is a picture of $O(2, 2)$. The $()$-shaped space between the spheres is not part of the surface.
A similar space with the copies of $D^1$ replaced by copies of $D^1 \times [0, 1]$, identifying the spheres at the ends of the interval, is homotopy equivalent to $O(2, 2)$.
These long discs can then be pinched down to copies of the interval $[0, 1]$, and the space in the neighbourhood of the intersection of the sphere pushed around to compensate for the shrinking.
Finally, we can retract one interval to a point, which we will take to be the basepoint for the wedge. We slide the endpoints of the other interval to the basepoint, which gives us a wedge summand of a circle.