I am trying to see why is it that if we have a transitive action of a lie group $G\rightarrow M$ then if this action is transitive the connected component of the identity $G^0$ also acts transitively on $M$ and that for all $p\in M$, $G/G_0 \cong G_p/(G_p\cap G^0)$.
Since the action is transitive we know that $G/G_p\cong M$ and since the map $G\rightarrow G_p$ is a submersion we get that the map $G\rightarrow M, g\rightarrow g.p$ is open . I don't know how useful this is since we only know that $G^0$ is a closed set I don't think it has to be open since connected components don't necessarily need to be open, and also I am not sure if this map is even closed or if it is how I could try and prove it.
Any enlightment is appreciated. Thanks in advance.
Manifolds (and thus also Lie groups) inherit the property of being locally connected from Euclidean space, and connected components of locally connected spaces are both closed and open. Hence $G_0$ is open in every Lie group.
Further, if we know that for every $q \in M$ the map $\phi_q : G \to M, g \mapsto g \cdot q$ is open, then the orbit $\phi_q(G_0) \subset M$ must be open. Then, fixing a single $p \in M$, the union $\bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$ is open as a union of open sets. But note that $$M = \bigcup_{q \in M} \phi_q(G_0)= \phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0),$$ because every $m \in M$ is equal to $e \cdot m \in \phi_m(G_0)$ if $e \in G_0$ is the identity.
But then we have written the connected set $M$ as a disjoint union of open sets $\phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$, and since the first set is nonempty, this must mean that the second one is empty, and hence $\phi_p(G_0) = M$.
(For academic integrity: Pieced this answer together from the comments to the very similar question here.)