Explicitly, what are the maps (and what are the spaces) in the BGG analysis of the complete flag manifold of $\operatorname{SL}(3,\mathbb{C})$? I will be referring to Baston--Eastwood for this question.
In Chapter 8 of Baston--Eastwood, I believe that I now understand why the crossed Dynkin diagram with labels $(2,-1) \oplus (-1,2)$ correspond to the matrix elements
$$\begin{pmatrix}
0 & 0 & 0 \\
\ast & 0 & 0 \\
0 & \ast & 0
\end{pmatrix}\,.$$
These correspond to certain generators in the $\mathfrak{sl}(3,\mathbb{C})$ algebra, and thus can be viewed as tangent vectors in the tangent space at the identity coset of $\operatorname{SL}(3,\mathbb{C})/B$.
On page 82, the non-integrability of the associated distribution seems to imply that there is an exact sequence
$$0 \rightarrow \mathbb{C} \rightarrow (0,0) \rightarrow (-2,1) \oplus (1,-2)\,.$$
It is my understanding that the representations associated to $(-2,1) \oplus (1,-2)$ corresponds to the cotangent space at the identity coset. This might be my first mistake, but I'd like confirmation that this either is or is not the case.
The next statement in the book is about the kernel of the projection map $\Omega^1 \rightarrow \Delta^1$. It's not clear to me what this map is, nor what the space $\Delta^1$ is. My first inclination was that the map was the $1$-form Laplacian, and the space $\Delta^1$ is those forms that can be written as $\Delta \omega$, for $\omega \in \Omega^1$---but this seems incorrect, especially because it seems that $\Delta^1 \cong (-2,1) \oplus (1,-2)$---but the cotangent space should be the same as the space of $1$-forms.
My confusion continues elsewhere: on page 84, $\omega_3 \in (-1,-1)$ is treated as a $1$-form, but it's also treated as a scalar! How can that be, unless there is some implied isomorphism?
I'm sure I'm simply missing some connection here that will make all of these operations and spaces make sense: my hope is that someone more familiar with this material (and this book in particular) will be able to enlighten me.
 A: So $D = (2,-1) \oplus (-1,2)$ is not the whole tangent space but a rank 2 (non-integrable) distribution inside it. Similarly $\Delta^1$ is the dual space to $D$ so $\Delta_1 =(-2,1) \oplus (1,-2)$ (which is clear from natural observations about parabolic subalgebras) is not the whole cotangent space but only a 2-dimensional subspace. Then $(-1,-1)$ completes the cotangent space so it is a natural kernel of the projection $\Omega^1 \to \Delta^1$. We should perhaps be more careful about what is a quotient and what is a subspace but I think there are assumed to be a bunch of identifications floating around.
So to conclude (sections of) $(-1,-1)$ are just a specific type of 1-form. I don't see a place where they treat it as a scalar (my page numbers are definitely not the same as yours I guess we have different editions). I see somewhere where they find a map from $(-1,-1)$ to itself and show that must be a non-zero scalar and a place where they differentiate $\omega_3$ and feed it two vector fields which outputs a number but that is just how $2$-forms work.
A: Firstly, for concreteness, $B$ is the (Borel) subgroup of upper triangular matrices in $\mathrm{SL}(3, \Bbb C)$.
(1) The images of the elements of $\mathfrak{sl}(3, \Bbb C)$ of the form $$\pmatrix{\cdot&\cdot&\cdot\\\ast&\cdot&\cdot\\\cdot&\ast&\cdot}$$ under the projection $\mathfrak{sl}(3, \Bbb C) \to \mathfrak{sl}(3, \Bbb C) / \mathfrak{b}$ comprise a $2$-dimensional subspace of $\mathfrak{sl}(3, \Bbb C) / \mathfrak{b} \cong T_{I \cdot B} (\operatorname{SL}(3, \Bbb C) / B)$, and not the full tangent space. Critically, this subspace is a $B$-module. Extending this subspace by left-invariance defines a $2$-plane distribution $D$ on $\operatorname{SL}(3, \Bbb C) / B$, and the form of the Lie bracket shows that $D$ is a contact distribution, i.e. that $$[D, D] = T := T(SL(3, \Bbb C) / B) .$$
Now, the dual $\Delta^1 := D^*$ of $D$ is the space of $1$-forms on $D$. The map dual to the inclusion $D \hookrightarrow T$ is the surjection $$\Omega^1 := T^* \to D^* =: \Delta^1$$ given by restricting the domain of an element of, say, $T_p^*$, viewed as a map $\Omega_p \to \Bbb C$, to $D_p$.
By definition the map $\delta$ is just the composition of the exterior derivative $d$ followed by the projection onto $\Delta^1$, i.e., take the exterior derivative and then forget the values of that derivative everywhere except on the distribution $D$:
$$\Delta^0 \stackrel{d}{\to} \Omega^1 \to \Delta^1.$$
The fact that $D$ is contact guarantees that $$0 \to \Bbb C \to \Delta^0 \stackrel{\delta}{\to} \Delta^1$$ is exact (here, $\Delta^0 = \Omega^0$ is the bundle whose sections are functions $\mathrm{SL}(3, \Bbb C) / B \to \Bbb C$). Informally, $\delta$ is an analogue of the exterior derivative $d$ that takes into account the structure $D$.
(2) Later on, the text defines $\omega_3 \in \Omega^1$ to be the left-invariant $1$-form on $T$ characterized by $$\omega_3: \pmatrix{\ast&\ast&\ast\\\ast&\ast&\ast\\ Y_3&\ast&\ast} \mapsto Y_3$$ on $T_I \cong \mathfrak{sl}(3, \Bbb C)$. Since $\omega_3$ annihilates $D$, we can also view $\omega_3$ as a map (in fact, isomorphism) $T / D \to \Bbb R$, or equivalently as an element of the bundle $(T / D)^* = (-1, -1)$. The dual of the canonical projection $T \to T / D$ is the inclusion $(T / D)^* \to \Omega^1$ that regards an element of $(T / D)^*$ as a $1$-form that annihilates $D$. Elements of $(T / D)^*$ are not scalars in the strict sense but they are elements of a trivial line bundle.
The above algebraic maps fit together into a short exact sequence of vector bundles,
$$0 \to D \to T \to T / D \to 0,$$
and its dual,
$$0 \to (T / D)^* \to \Omega^1 \to \Delta^1 \to 0 .$$
Remark As an aside, the sets of elements $$\pmatrix{\cdot&\cdot&\cdot\\\ast&\cdot&\cdot\\\cdot&\cdot&\cdot} \qquad \textrm{and} \qquad \pmatrix{\cdot&\cdot&\cdot\\\cdot&\cdot&\cdot\\\cdot&\ast&\cdot}$$ each project to $1$-dimensional subspaces of $\mathfrak{sl}(3, \Bbb C) / \mathfrak{b}$, so the plane field $D$ in fact inherits additional structure for free, namely, a direct sum decomposition $D = E \oplus F$ into line fields $E, F$. These line fields correspond (separately) to the summands $(+2, -1)$ and $(-1, +2)$. Note that $E \otimes F = (+2, -1) \otimes (-1, +2) \cong (+1, +1) = TM / D$.
