In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, ie about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$.

I don't really understand what you mean more generally about the group structure on $Emb(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by `connected sum', there doesn't seem to be an inverse with respect to connected sum of $2$-spheres. It's different in Habegger's situation, because the codimension is bigger than two.

In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, i.e. about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$.

I don't really understand what you mean more generally about the group structure on $\operatorname{Emb}(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by ‘connected sum’, there doesn't seem to be an inverse with respect to connected sum of $2$-spheres. It's different in Habegger's situation, because the codimension is bigger than two.

In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, ie about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$.

I don't really understand what you mean more generally about the group structure on $Emb(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by connected sum, there doesn't seem to be an inverse with respect to connected sum of $2$-spheres.

In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, ie about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$.

I don't really understand what you mean more generally about the group structure on $Emb(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by `connected sum', there doesn't seem to be an inverse with respect to connected sum of $2$-spheres. It's different in Habegger's situation, because the codimension is bigger than two.