In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.

Your example is of a very special type, where $G$ is a subgroup of a reductive group $H$, so instead of worrying directly about the nonreductive GIT quotient $X//G$ we can instead look at $(X \times H//G)//H$ and punt the nonreductivity issue to the more-universal problem of computing $H//G$. In the case at hand, $H = SL_2$, and $H//G \cong \mathbb A^2$. (Note that the map $H \to H//G$ is not onto!)

Instead of taking $X = \mathbb P^{2r+1}$, let me keep scaling in abeyance, to quotient by later, taking $X = \mathbb A^{2r+2}$. Then $(X \times H//G)//H$ is the affine cone over $Gr(2,r+1)$. What's a little tricky, then, is that this remaining $\mathbb G_m$ action is not acting on said cone by dilation, because it doesn't act on the factor of $\mathbb A^2$ we just attached. Instead we get one of the "weighted Grassmannians" of Corti and Reid. I don't have a more detailed answer than that, but at least this replaces your original $G$-quotient question by a $\mathbb G_m$-quotient question.

In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.

Your example is of a very special type, where the action of $G$ can be extended to that of a reductive group $H \geq G$, so instead of worrying directly about the nonreductive GIT quotient $X//G$ we can instead look at $(X \times H//G)//H$ and punt the nonreductivity issue to the more-universal problem of computing $H//G$. In the case at hand, $H = SL_2$, and $H//G \cong \mathbb A^2$. (Note that the map $H \to H//G$ is not onto!)

Instead of taking $X = \mathbb P^{2r+1}$, let me keep scaling in abeyance to quotient by later, and take $X = \mathbb A^{2r+2}$. Then $(X \times H//G)//H$ is the affine cone over $Gr(2,r+1)$. What's a little tricky, then, is that this remaining $\mathbb G_m$ action is not acting on said cone by dilation, because it doesn't act on the factor of $\mathbb A^2$ we just attached. Instead we get one of the "weighted Grassmannians" of Corti and Reid. I don't have a more detailed answer than that, but at least this replaces your original $G$-quotient question by a $\mathbb G_m$-quotient question.

In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.

Your example is of a very special type, where $G$ is a subgroup of a reductive group $H$, so instead of worrying directly about the nonreductive GIT quotient $X//G$ we can instead look at $(X \times H//G)//G$ and punt the nonreductivity issue to the more-universal problem of computing $H//G$. In the case at hand, $H = SL_2$, and $H//G \cong \mathbb A^2$.

Instead of taking $X = \mathbb P^{2r+1}$, let me keep scaling in abeyance, to quotient by later, taking $X = \mathbb A^{2r+2}$. Then $(X \times H//G)//H$ is the affine cone over $Gr(2,r+1)$. What's a little tricky, then, is that this remaining $\mathbb G_m$ action is not acting on said cone by dilation, because it doesn't act on the factor of $\mathbb A^2$ we just attached. Instead we get one of the "weighted Grassmannians" of Corti and Reid. I don't have a more detailed answer than that, but at least this replaces your original $G$-quotient question by a $\mathbb G_m$-quotient question.

In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.

Your example is of a very special type, where $G$ is a subgroup of a reductive group $H$, so instead of worrying directly about the nonreductive GIT quotient $X//G$ we can instead look at $(X \times H//G)//H$ and punt the nonreductivity issue to the more-universal problem of computing $H//G$. In the case at hand, $H = SL_2$, and $H//G \cong \mathbb A^2$. (Note that the map $H \to H//G$ is not onto!)

Instead of taking $X = \mathbb P^{2r+1}$, let me keep scaling in abeyance, to quotient by later, taking $X = \mathbb A^{2r+2}$. Then $(X \times H//G)//H$ is the affine cone over $Gr(2,r+1)$. What's a little tricky, then, is that this remaining $\mathbb G_m$ action is not acting on said cone by dilation, because it doesn't act on the factor of $\mathbb A^2$ we just attached. Instead we get one of the "weighted Grassmannians" of Corti and Reid. I don't have a more detailed answer than that, but at least this replaces your original $G$-quotient question by a $\mathbb G_m$-quotient question.