Why does the projection $\pi:(a,b,c) \in S^2 \to[a,b,c] \in P^2$ have rank 2 everywhere?

$\pi$ is a local diffeomorphism which shows that it has rank $2$ everywhere. In fact, the six sets $$U_j^\pm = \{(x_1,x_2,x_3) \in S^2 \mid (-1)^{\pm 1} x_j > 0 \}$$ with $j = 1,2,3$ are open and cover $S^2$. They are mapped by $\pi$ diffeomorphically onto the three open sets $$V_j = \{ [x_1,x_2,x_3] \mid x_j \ne 0\} \subset \mathbb P^2.$$

$g$ is a immersion at $p$ iff $dg_{[p]}$ has maximal rank (i.e. rank $2$). Write $h = G \mid_{S^2}$. We have $dh_p = dg_{[p]}\circ d\pi_p$. (I prefer to write $dh_p$ instead of $h_{*p}$; this notation is somewhat unusual.) Since $d\pi_p$ is an isomorphism, the rank of $dh_p$ agrees with the rank of $dg_{[p]}$ and it suffices to determine all $p$ such that $dh_p$ has rank $2$. This means that it maps no non-zero vector of $T_pS^2$ to $0$. Therefore $g$ is not an immersion at $[p]$ iff there exist a non-zero $(p,\nu)\in T_pS^2$and such that $dh_p(p,\nu)=0$.