Quotient of $GL_n$ by unipotent group and maximal torus Assume we are working over a field $k$ and $G= GL_n$. Then we get the (standard) maximal torus $T$, the unipotent group $U$ and the Borel $B$ of upper triangular matrices. More generally we have we can choose a parabolic $P$ with its Levi $M$.
The quotient $G/P$ admits a nice description and is a projective variety. For example
$$ G/B(R)=\{\Lambda_1\subset\Lambda_2\subset...\subset \Lambda_n=R^n| \Lambda_i \textrm{ is projective of rank } i \textrm{ and the inclusions are direct summands}\}. $$
If $R$ is a field, then this is just the set of full flags.
My question primarily is: What does $G/U$ look like?
I suspect that $G/U\to G/B$ is a $T$-torsor (resp. $M$-torsor in the parabolic case) and that it should admit a moduli description, something like
$$ G/U(R)=\{\Lambda_1\subset\Lambda_2\subset...\subset \Lambda_n=R^n| \Lambda_i \textrm{ as above }+ \textrm{ an automorphism } g:\Lambda_{i+1}/\Lambda_i\to\Lambda_{i+1}/\Lambda_i \}. $$
Or maybe a condition on splittings $\Lambda_{i+1}\to\Lambda_i$?
A second question is:  Can we do something like this for $G/T$ as well?

According to this answer, working over $\mathbb{C}$, $GL_2(\mathbb{C})/U(\mathbb{C})\cong \mathbb{C}^2-\{0\}$, but I think this is wrong: $U(\mathbb{C})$ is not the stabilizer of $v=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, as e.g. $ \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix} $ also stabilizes $v$.
As noted in the comments, in the linked answer the group in question is $SL_2$ and not $GL_2$, so my specific objection here is obsolete
 A: Question: "According to this answer, working over $\mathbb{C}, GL_2(\mathbb{C})/U(\mathbb{C})≅\mathbb{C}^2−\{0\}$, but I think this is wrong: $U(\mathbb{C})$ is not the stabilizer of $v=(1,0)$, as e.g. $(1,0,0,λ)$ also stabilizes v."
Answer: If $V:=\mathbb{C}^2$, $U \subseteq B$ it follows $SL(V)/B \cong \mathbb{P}^1$, and the map
$$\pi: SL(V)/U \rightarrow SL(V)/B$$
is the canonical map $\pi: Y:=\mathbb{C}^2-\{0\} \rightarrow \mathbb{P}^1$ which is a "quotient map" for the canonical $\mathbb{C}^*$-action on $Y$. Let $v:=(1,0)\in V$ and define a map
$$\rho: SL(V) \rightarrow V-\{(0,0)\}$$
by $\rho(g):=gv$. It follows the stabilizer sub-group of $SL(V)$ of $v$ is $U$, and the subgroup of elements $g\in SL(V)$ with $gv \in \mathbb{C}v$ is the group $B$. There are isomorphisms $SL(V)/U \cong \mathbb{C}^2-\{(0,0)\}$ and $SL(V)/B \cong \mathbb{P}^1$. The map $\pi$ is a principal $\mathbb{C}^*$-bundle on $\mathbb{P}^1$.  There is an action of $\mathbb{C}^*$ on $Y$ and an isomorphism $Y/\mathbb{C}^* \cong \mathbb{P}^1$.
Similar constructions exist for the general case $G/U \rightarrow G/B$. In general for projective space you may construct $\mathbb{P}^n$ as a quotient
$$\pi:\mathbb{A}^{n+1}_{\mathbb{C}}-\{0\} \rightarrow \mathbb{P}^n \cong \mathbb{A}^{n+1}_{\mathbb{C}}-\{0\}/\mathbb{C}^*:=Z/\mathbb{C}^*$$
and you can construct projective space in two ways: As $SL(n+1,\mathbb{C})/B$ or as $Z/\mathbb{C}^*$. The map $\pi: Z \rightarrow SL(n+2, \mathbb{C})/B$ is a principal $\mathbb{C}^*$-bundle.
If $v\in V$ and $B$ is the subgroup of $SL(V)$ sending $v$ to $\mathbb{C}v$ and $U$ is the subgroup fixing $v$ it follows $U \subseteq B$ is a normal subgroup. The fiber of the projection map $\pi$ at the point $[v]\in \mathbb{P}(V^*)$ is the group $B/U$: If $\overline{x} \in B/U$ it follows
$$\pi(\overline{x}):=[xv]=[v].$$
Moreover if $\pi(\overline{g})=[v]$ it follows $\overline{g}\in B/U$.
Question: "My question primarily is: What does G/U look like?"
The subgrop $U \subseteq B$ is a normal subgroup, the map $\pi: G/U \rightarrow G/B$ has fiber $B/U$: There is an open cover $U_i$ of $G/B$ and an isomorphism $\pi^{-1}(U_i) \cong U_i \times B/U$.
Note: In general if $H \subseteq G$ are linear algebraic groups over $K:=\mathbb{C}$ you may construct the quotient $\pi: G \rightarrow G/H$ and $H$ is a smooth quasi projective variety over $K$. If $G$ is semi simple and $H$ is a parabolic sub-group, it follows $\pi$ is locally trivial in the Zariski topology. There is a Zariski open cover $U_i$ of $G/H$ and isomorphisms $\pi^{-1}(U_i) \cong U_i \times H$. Hence $\pi$ is a principal fiber bundle over $G/H$ with fiber $H$.
Note also that if $G$ is semi simple and $P \subseteq G$ is a parabolic subgroup, there is a finite dimensional irreducible $G$-module $V$ and a non-zero vector $v\in V$, where $P$ is the stabilizer group of $v$: $P$ is the subgroup of $G$ of elements $g$ with $g(v) \in \mathbb{C}v$.
Note also that if $G:=GL(V)$ any flag variety $F:=G/P$ may be constructed as $F \cong SL(V)/P'$ for some parabolic $P' \subseteq SL(V)$. Any linear algebraic group $H$ over $K$ is a closed subgroup of $GL(V)$ for some finite dimensional $V$.
Note: If you consider the grassmannian $G(k,E)$ and flag bundle $F(d,E)$ over a scheme $S$: Maybe SGA3 or Jantzen's book.
Note: I want to warn you that the "functor of points" point of view of grothendieck requires knowledge on "universes", "inaccessible cardinals" and lots of "category theory". Hence you have to consume much non-standard set theory.
