Some comments.
The space of sections of $LX \to X$ is a grouplike topological monoid, by pointwise multiplication. So, its path components do form a group. In fact it is a loop space $\Omega_1F(X,X)$ of the space of functions $F(X,X)$, where loops are based at the identity map $1\!\!\! : X \to X$.
For $A\to X$ a map, let $\Gamma(LX|A)$ denote the space of sections of $LX \to X$ along $A$. This is the same as the space of sections of the pullback $A \times_X LX \to A$. This is also a loop space, this time of $F(A,X)$.
When $A = X$ we set $\Gamma(LX) = \Gamma(LX|A)$.
The restriction map $$ F(X,X) \to F(A,X) $$ is a fibration and it loops to the restriction map $\Gamma(LX) \to \Gamma(LX|A)$ . Its fiber at the constant map $A\to X$ is the based function space $F_*(X/A,X)$.
Let's apply the above to the inclusion $\ast \subset S^2$ basepoint. This gives a homotopy fiber sequence $$ \Omega^2 S^2 \to F(S^2,S^2) \to S^2\, , $$ where the second map is given by evaluation.
Take $\pi_*$ (with respect to the correct basepoint "1") to get an exact sequence of groups $$ \cdots \to \pi_2(S^2) \to \pi_1(\Omega^2S^2;1) \to \pi_1(F(S^2,S^2);1) \to 1 $$ where $\pi_1(F(S^2,S^2);1) = \pi_0(LS^2)$, and $\pi_2(S^2) = \Bbb Z = \pi_1(\Omega^2S^2;1)$. It follows that there's a short exact sequence of groups $$ \Bbb Z \to \Bbb Z \to \pi_0(LS^2) \to 1\, . $$ It therefore suffices to compute the homomorphism $\Bbb Z \to \Bbb Z$. To compute this, you'll need to understand how the boundary operator works. This amounts in the end to the computation discussed by "user95545" above.