(Note: $O(n) = O(n, \Bbb{R})$)
I've notice that $O(1)$ is equivalent to $S^0$. And I've read that $O(2)$ is equivalent to two copies of the circle group $S^1$.
I was wondering if someone can explain why this is?
Also is there a general (similar) rule for O(n)?
I don't know what you mean with “equivalent”. However, $O(1,\mathbb{R})=\{1,-1\}=S^0$. On the other hand$$O(2,\mathbb{R})=\left\{\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\,\middle|\,\theta\in\mathbb R\right\}\cup\left\{\begin{pmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{pmatrix}\,\middle|\,\theta\in\mathbb R\right\}$$and this expresses $O(2,\mathbb{R})$ as the union of two disjoint sets, each of which is homeomorphic with $S^1$.
On the other hand, $O(3,\mathbb R)$ is $3$-dimensional, whereas $S^2$ is $2$-dimensional. Therefore, there is no similar relation between $O(3,\mathbb{R})$ and $S^2$.