Question 1 : The definition of the tangent/cotangent bundle is the usual definition used in differential geometry, see wikipedia for example. However, if $G$ is a Lie group it is possible to show that there is an isomorphism of vector bundle $TG \cong G \times \mathfrak g$. This isomorphism uses the group structure, it is not true in general that $TM \cong M \times \Bbb R^m$ (where $m = \dim M$), for example $M=S^2 = \Bbb P^1$.
The other part of question $1$ was answered in comments.
Question 2 : Describing explicitly $T^* \mathcal B$ is precisely what is done in Chriss-Ginzburg, where they give several different descriptions.
$\mathcal B$ is defined as the set of Borel subalgebra of $\mathfrak g$, and it is a classical result recalled in Chriss-Ginzburg that all Borel subalgebras are conjugated. Hence $G$ acts transitively by conjugaison action on $\mathcal B$. Moreover, if we fix a special Borel $B$, then $B = \mathrm{Stab}_G(\mathfrak b)$ (since Borel are self-normalizing) which implies $\mathcal B \cong G/B$.
Question 3 : One has $\mathcal F \cong \mathcal B$ because for example they both correspond to $G/B$, and this gives the structure of a smooth compact manifold. Alternatively, you can embedd $\mathcal F$ in a product of grassmannians.
Proposition 4.1.2 is an alternative description of $T^*(\mathcal B)$, so you should try to see it as a different description of the same object, rather as a similar description. For example it is not obvious at all that $M$ has the structure of smooth variety, let alone a vector bundle over $\mathcal B$ !
Firstly, yes (conjugacy classes of) parabolic subalgebras are in 1-to-1 correspondence with subsets of the simple roots as you say and we denote these with crossed nodes on our Dynkin (or Satake) diagrams.
Secondly, irreducible representations are in correspondence with dominant integral weights (i.e. their highest weights). In other words these are weights which are a linear combinations of the fundamental weights $\omega_i$ (with coefficients $a_i$ that are nonnegative integers) which are natural paired with the simple roots. Specifically, they are the dual basis to the simple coroots. Thus we can specify an irreducible representation again by a decorated Dynkin diagram where we put $a_i$ over the corresponding node.
Baston and Eastwood go on to look at how the parabolic subalgebra acts on representations of $\mathfrak{g}$. Now there are a few nice interactions of this notation. If we take a given parabolic subalgebra and choose a representation $V$ which has $0$'s on the uncrossed nodes and non-zero integers on the crossed nodes, the parabolic subalgebra induces a filtration (or flag) on the representation which terminates in a 1-dimensional subspace (i.e. a line). Now looking at the conjugacy class of this parabolic we get a manifold $G/P$ and with each element we associate a line in $V$. In fact this is a realisation of $G/P$ as a projective variety.
More generally, for an arbitrary representation $V$ we just get a flag in $V$. Hence we call $G/P$ a flag manifold (or variety).
The smallest subspace in the flag is an irreducible representation of $\mathfrak{p}$ which they denote with a Dynkin diagram with crossed nodes and numbers. [Edit: this only accounts for representations where all the numbers are positive integers, I will have a think about this more carefully and come back to fix it]. They then consider the space $\mathfrak{g}/\mathfrak{p}$ which can be identified with the tangent space $T_{\mathfrak{p}}G/P$. If your conception of $\mathfrak{p}$ is as the upper triangular matrices (so it is Borel) then you could realise this as the strictly lower triangular ones. The filtration induced on $\mathfrak{g}$ looks like: $$\mathfrak{g} = \mathfrak{p}^{(2)} \geq \mathfrak{p}^{(1)} \geq \mathfrak{p}^{(0)} \geq \mathfrak{p}^{(-1)} \geq \mathfrak{p}^{(-2)} \geq \{0\}$$
Where $$ \mathfrak{p}^{(1)} = \begin{pmatrix} *&*&*\\ *&*&* \\0&*&* \end{pmatrix}, \mathfrak{p}^{(0)} = \mathfrak{p} = \begin{pmatrix} *&*&*\\ 0&*&* \\0&0&* \end{pmatrix}, \mathfrak{p}^{(-1)} = \begin{pmatrix} 0&*&*\\ 0&0&* \\0&0&0 \end{pmatrix}, \mathfrak{p}^{(-2)} = \begin{pmatrix} 0&0&*\\ 0&0&0 \\0&0&0 \end{pmatrix}$$
Now look at $\mathfrak{p}^{(1)}/ \mathfrak{p} \leq \mathfrak{g}/\mathfrak{p}$ which can be identified with $\begin{pmatrix} 0&0&0\\ *&0&0 \\0&*&0 \end{pmatrix}$ and ask how does $\mathfrak{p}$ act on it. The nilradical $\mathfrak{p}^{(-1)}$ acts trivially so this is really just an action by the Levi subalgebra which in this case is exactly the Cartan subalgebra. Comparing this action with the irreducible $\mathfrak{p}$ representations we defined a minute ago and we see these are the ones defined by the given choices of $\mathfrak{g}$ representation.
Edit: I've taken a longer look at this and I'm still not sure exactly how to find the irreducible reps for weights that are only dominant for $\mathfrak{p}$ and not $\mathfrak{g}$. Even using example 3.2.2 has left me confused. They say there that our Dynkin diagram (two crosses) with labelling $(p,q)$ corresponds to the 1-dimensional representation $L^p \otimes (H/L)^{p+q}$ or $(H/L)^p \otimes (V/H)^{p+q}$ where I imagine powers mean tensor powers, $L^{-1}$ means $L^*$ and $\{0\} \leq L \leq H \leq V = \mathbb{C}^3$ is our flag. However, the example we are looking at is $(L^* \otimes H/L) \oplus ((H/L)^* \otimes V/H)$ both of which correspond to $(-1,2)$ apparently. $(2,-1)$ seems then to correspond to $L^2 \otimes H/L$ or $(H/L)^2 \otimes V/H$ which do not fit into $\mathfrak{g} \leq V^* \otimes V$. This all seems to contradict the story you identify later in the book so either I've seriously misunderstood something here or there's a mistake in one of these parts.
The one thing I have gleaned from the examples is that we can think of the numbers over the crosses can be thought of as taking tensor powers of the "factors" of our representations.
Edit 2: Okay I've thought about this even more. What is happening in this case should go as follows. Since we have picked a Borel subalgebra, every weight (in any representation) is fixed by the Levi since it is just a Cartan subalgebra. Thus the representation specified by $(p,q)$ over 2 crossed nodes is just the weight space for the weight $p\omega_1 + q\omega_2$ thought of as a 1-dimensional representation over the Levi. You may pick your favourite representation containing that weight for this to live inside. In your scenario we note that those two weights are actually roots so we can think of them inside the adjoint representation. I still think there is something wrong with example 3.2.2 or at least my understanding of it.
For the general case of a partial flag manifold I believe the story would go instead: now that the Levi is no longer a Cartan subalgebra it can generate bigger irreducible representations. In order to avoid over counting we need to find the highest weight for these and that leads back to the idea that the uncrossed nodes must have positive numbers attached. Again, the representation could be found inside some $\mathfrak{g}$ representation containing that weight (we may need to choose one that has a 1-dimensional weight space for that weight). Then the representation is simply that generated by the action of the Levi on this chosen weight space.
Best Answer
This is not a fiber product. It's why I think a better notation is $G \times^B \mathfrak n$. Also be careful with the notations : usually $B$ is a fixed Borel (in $G/B$ for example) and sometimes it may vary. So here we fix a Borel subgroup $B$.
Anyway, the construction is as follows : if $H \subset G$ is a closed subgroup and $E$ is a $H$-module, we can construct a vector bundle over $G/H$ given by $G \times^H E := (G \times E)/H$. Here $H$ acts by $h \cdot (g,e) = (gh, \rho(h^{-1})e)$. The construction is detailed here.
You can look at the map $\beta : G \times^B \mathfrak n \to \mathcal N'$, sending $(gB,x) \mapsto (Ad_g(x),gB)$, where $Ad_g(x)$ is the adjoint action, i.e $gxg^{-1}$ for matrix groups.
Now the claim follows if we can show that the map $G \times^B \mathfrak n \to \mathcal N'$ is an isomorphism (because we saw that $G \times^B \mathfrak n$ is a naturally a vector bundle over $\mathcal B$.) I think it is proved in Chriss-Ginzburg, but let us give the proof.
The maps is clearly surjective, since for any $x \in \mathfrak b'$ nilpotent, we have $\beta(g,Ad_g^{-1}(x)) = (x, \mathfrak b')$, where $gB \in \mathcal B$ correspond to $\mathfrak b'$.
Suppose $\beta(g,x) = \beta(g', x')$. Then, $gB = g'B$, so there is $b \in B$, such that $g = g'b$. From this and the equality $Ad_g(x) = Ad_{g'}(x')$ it is easy to conclude that $(g,x) = (g',x')$ in $G \times^B \mathfrak n$.
EDIT : In fact, the map $\beta$ was already constructed before in Corollary 3.1.33, to show the isomorphism $G \times^B \mathfrak b \cong \widetilde{\mathfrak g}$. Hence, to show that $G \times^B \mathfrak n \to \mathcal N'$, you just need to show surjectivity, which follows from the fact that all Borel are conjugated.