$\mathbb{P}^2(\mathbb{R})/H$ identifies the topological quotient space given by the relation $x \sim y \Leftrightarrow x=y$ or $x,y \in H$.
I tried to build an identification $\mathbb{P}^2(\mathbb{R}) \longrightarrow S^2$ defined by $[v] \longmapsto \frac{v}{||v||}$, but it doesn't seem to work.
Let $p : \mathbb R^3 \setminus \{0\} \to \mathbb P^2 = \mathbb P(\mathbb R^2)$ denote the quotient map.
A projective line $H \subset \mathbb P^2$ is given as $p(H' \setminus \{0\})$, where $H'$ is a two-dimensional linear subspace of $\mathbb R^3$. Equivlently, a subset $H \subset \mathbb P^2$ is a projective line if $H' = p^{-1}(H) \cup \{0\}$ is a two-dimensional linear subspace of $\mathbb R^3$.
It is easy see that $\mathbb P^2 /H_1 \approx \mathbb P^2 /H_2$ for any two projective lines $H_i$. In fact, let $\phi'$ be a linear automorphism on $\mathbb R^3$ such that $\phi'(H'_1) = H'_2$. Note that $\phi$ is a homeomorphism. Clearly $\phi'$ induces a homeomorphism $\phi$ on $\mathbb P^2$ such that $\phi(H_1) = H_2$. Hence we get a homeomorphism $\bar \phi : \mathbb P^2/H_1 \to \mathbb P^2/H_2$.
It therefore suffices to show that $\mathbb P^2/H_0 \approx S^2$, where $H_0$ is the special projective line such that $H_0' = \overline{\mathbb R}^2 = \{(x,y,0) \mid x,y \in \mathbb R\}$. Let $S^2_+ = \{(x,y,z) \in S^2 \mid z \ge 0\}$ denote the closed upper hemisphere of $S^2$. It is well-known that $$\mathbb P^2 = S^2_+ / \sim $$ where $\xi \sim \eta$ iff $\eta = \pm \xi$. Note that the only pairs of distinct points identified by $\sim$ lie on $\bar S^1= \{(x,y,0) \mid (x,y) \in S^1\}$. Under the identification $\mathbb P^2 = S^2_+ / \sim $ the projective line $H_0$ corresponds to $\bar S^1/\sim$ (which is homeomorphic to $S^1$ although we do not need to know this). Therefore $$\mathbb P^2 /H_0 \approx (S^2_+/\sim)/(\bar S^1/\sim) \approx S^2_+/\bar S^1 \approx S^2 .$$