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The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

e: $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ is the quotient of $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$ by the involution given by precomposing with the antipodal map. The fixed point set is of infinite codimension. Its complement, then, by transversality arguments, has the same fundamental group; and away from the fixed point set this is a covering map. So we obtain that $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ has fundamental group $\Bbb Z$, and the quotient map induces multiplication by two on the fundamental group.

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

e: $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ is the quotient of $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$ by the involution given by precomposing with the antipodal map. The fixed point set is of infinite codimension. Its complement, then, by transversality arguments, has the same fundamental group; and away from the fixed point set this is a covering map. So we obtain that $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ has fundamental group $\Bbb Z$, and the quotient map induces multiplication by two on the fundamental group.

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

e: $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ is the quotient of $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$ by the involution given by precomposing with the antipodal map. The fixed point set is of infinite codimension. Its complement, then, by transversality arguments, has the same fundamental group; and away from the fixed point set this is a covering map. So we obtain that $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ has fundamental group $\Bbb Z$, and the quotient map induces multiplication by two on the fundamental group.

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective holomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)