The cotangent bundle on the flag variety This has been asked previously here, though the only similarity between that question and mine is their question 1 (all put below). Moreover, the top answer of that post deals with $\mathfrak{g}$ from a geometric perspective - I am interested in the other way. I also have several other questions not addressed in that post.
To begin with, Lemma 1.4.9 of Chriss and Ginzburg's "Complex Geometry and Representation Theory", re-worded to emphasise what I care about:

There is a natural vector bundle isomorphism $T^*(G/P) \cong G \times_P \mathfrak{p}^\perp$.

Question 1: I am not sure how to even explicitly describe the cotangent bundle $T^*(G/P)$. Here, $P$ is a Lie subgroup of $G$. Obviously in the text, $T_e(G/P) = \mathfrak{g}/\mathfrak{p}$, and similarly for $T_e^*(G/P) = \mathfrak{p}^\perp$. But is this equality the definition, or does it come as a result of the definition? And if the latter, what is the definition of the cotangent bundle here?
This result and the following Proposition 1.4.11 are used in the proof of Lemma 3.2.2.

There is a natural vector bundle isomorphism $\widetilde{\mathcal{N}} \cong T^*\mathcal{B}$.

Question 2: I am also not sure how to explicitly describe the cotangent bundle $T^*\mathcal{B}$. Here $\mathcal{B}$ is the flag variety, i.e. the set of all Borel subalgebras $ \mathfrak{b} \subset \mathfrak{g}$.
The proof of this lemma then proceeds by using the previous result to show that $T^*\mathcal{B} = G \times_\mathcal{B} \mathfrak{b}^\perp$. I understand that there is a bijection $G/B \cong \mathcal{B}$, though admit I am not familiar with this, only that we map $g \mapsto g \cdot \mathfrak{b} \cdot g^{-1}$. Clarification here would be appreciated, though is secondary to my other questions.
Finally, in Proposition 4.1.2:

There is a natural vector bundle isomorphism $T^* \mathcal{F} \cong M$, where $M = \{(x,F) \in \mathcal{N} \times \mathcal{F} : xF_i \subset F_{i-1} \forall i\}$.

Question 3: The proof with $\mathcal{F}$ the flag variety is supposedly analogous to the proof of the previous result, however I don't see what the analogue of $G \times_\mathcal{B} \mathfrak{b}^\perp$ is. The text mention that the set of partial flags $\mathcal{F}$ has a smooth compact manifold structure, but I am not familiar enough with the geometry perspective - what is this structure, and how do we get the Lie (sub)algebra structure from that? 
 A: 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 $\mathcal B$ !
