First look at it intuitively. The relation $(x,y,z) \sim_1 (tx,ty,tz)$ on $\mathbb{R}^3\setminus \{(0,0,0)\}$ collapses each line through the origin to a single point. Each of those lines through the origin intersects the sphere $S^2$ in exactly two points; if one of them is $(x,y,z)$, the other is $(-x,-y,-z)$, so the relation $(x,y,z) \sim_2 (-x,-y,-z)$ on $S^2$ collapses them to a single point as well. This point corresponds to the collapse of the line in the first quotient.
You can use this idea to find an actual homeomorphism. Let $q_1:\mathbb{R}^3\setminus \{(0,0,0)\}\to X$ be the first quotient map and $q_2:S^2\to Y$ be the second, where $X = \mathbb{R}^3\setminus \{(0,0,0)\}/\sim_1$ and $Y=S^2/\sim_2$. Let $p$ be any point of $Y$; then there is a point $(x,y,z)\in S^2$ such that $p = q_2((x,y,z))=q_2((-x,-y,-z)).$ Clearly $(x,y,z) \ne (0,0,0)$, so $q_1((x,y,z)) \in X$.
Now let $h(p) = q_1((x,y,z))$. Note that $q_1((-x,-y,-z))=q_1((x,y,z))$ (why?), so it doesn’t matter which of the two $q_2$-preimages of $p$ I use. I claim that $h$ is a homeomorphism from $Y$ onto $X$, but I’ll leave you to try to work out the details.