An exercise (8-12) in Lee's Introduction to Smooth Manifolds involves showing that if $F : \mathbb{R}^2 \to \mathbb{RP}^2$ is given by $F(x,y) = [x,y,1]$, then there is a vector field on $\mathbb{RP}^2$ that is $F$-related to the vector field $X = x\partial/\partial y - y\partial/\partial x$ on $\mathbb{R}^2$.
I solved this problem as follows: We begin by letting $U_1,U_2,U_3 \subset \mathbb{RP}^2$ be the open subsets on which the first, second, and third coordinates, respectively, are nonzero, and let $(u_i,v_i) : U_i \to \mathbb{R}^2$ be the usual coordinate systems for each $i = 1,2,3$. We then define a smooth vector field $Y_i$ in coordinates on each $U_i$ as follows: \begin{align*} Y_1 &= (u_1^2 + 1)\frac{\partial}{\partial u_1} + u_1v_1\frac{\partial}{\partial v_1} \\ Y_2 &= -(u_2^2 + 1)\frac{\partial}{\partial u_2} - u_2v_2\frac{\partial}{\partial v_2} \\ Y_3 &= -v_3\frac{\partial}{\partial u_3} + u_3\frac{\partial}{\partial v_3}. \end{align*} It's then a straightforward computation with Jacobians to show that these three vector fields agree on intersections, and so they extend to a smooth global vector field $Y$ on $\mathbb{RP}^2$. One more computation shows that $Y$ is $F$-related to $X$. (I might have made a computational error here but that's beside the point.)
Despite having a formula for the vector field $Y$, I still have no intuitive grasp of what it actually looks like. $\mathbb{RP}^2$ is already a pretty abstract object, and how to imagine vector fields on it is a mystery to me---the above coordinate representations don't shed that much light on its structure. Is there a coordinate-independent way to define $Y$? I'm thinking maybe we can define a visualizable vector field on $\mathbb{R}^3 \setminus \{0\}$ that sinks through the quotient map $q : \mathbb{R}^3 \setminus \{0\} \to \mathbb{RP}^2$, but I don't know how the details would work out.