I have read in several places that if $\Omega$ is the coadjoint orbit of $\zeta \in \mathfrak{g}^*$, the map from $G \to \Omega$ that sends $g \mapsto Ad^*(g)(\zeta)$ gives a surjection, and taking the differential of this map at the identity gives a surjection from $T_eG = \mathfrak{g} \to T_{\zeta}\Omega$, and then the $2$-form is defined by identifying elements of the tangent space with elements of $\mathfrak{g}$ and taking the Lie bracket.
But a priori this only seems to define the $2$-form at $\zeta$, ie:
$\omega(\zeta) (\hat{X}, \hat{Y}) = \zeta([X,Y])$
where $\hat{X},\hat{Y} \in T_\zeta \Omega$. But what about $\omega(\eta)$ for $\eta \in \Omega$ but $\omega \neq \zeta$? The surjection is from $\mathfrak{g} \to T_\zeta \Omega$, but not from $\mathfrak{g} \to T_{\eta}\Omega$.
The best way to understand these things is to cut through the fog of (mostly unnecessary) terminology by adopting a few conventions that will dramatically simplify everything and make it totally transparent. So, let's do that here, first.
A Lie group G is a manifold. As such, each element g ∈ G has a tangent space TgG. As is the case for all manifolds, if t ∈ ℝ ↦ g(t) ∈ G is a C1 curve describing a trajectory in G, then g'(t) ∈ Tg(t)G - which, in fact, is practically the way the tangent space is defined. Correspondingly, I will denote a generic element of TgG as g' below.
Collectively, tangent spaces inherit from the group a natural product on both the left and right g' ∈ TgG, g0,g1 ∈ G ↦ g0g'g1 ∈ Tg0gg1G; such that d/dt (g0 g(t) g1) = g0 g'(t) g1.
The Lie algebra L = TeG associated with G is the tangent space of the group's identity e ∈ G. Correspondingly one may define the adjoint action on L by Ad(g): e' ∈ L = TeG ↦ ge'g-1 ∈ Tgeg-1G = TeG = L.
But, as you can see, the obfuscation that mathematicians bury their formalism in hides a greater generalization, which they remain mostly oblivious to. For instance if g,h ∈ G with gh = hg, then you also have Ad(g): h' ∈ ThG ↦ gh'g-1 ∈ ThG, such that Ad(g)(he') = g(he')g-1 = ghe'g-1 = hge'g-1 = h Ad(g)e'; similarly Ad(g)(e'h) = Ad(g)e' h, for e' ∈ L. And you could just define Ad(g)h' = gh'g-1 over all the tangent spaces ThG, as long as you understand that it moves to a different tangent space Ad(g)h' ∈ Tghg-1G ≠ ThG if gh ≠ hg.
Taking another derivative yields a bracket operation such that [g'(t),h'] = d/dt (g(t)h'g(t)-1) ∈ ThG over h' ∈ ThG, provided that g(t)h = hg(t) for all t. It can be defined without restriction over L, since g(t)e = eg(t) already holds. Despite its greater transparency and greater directness, this is actually more general than the Lie bracket, in that the latter is restricted only to g' ∈ L, while I just defined e' ∈ L ↦ [g',e'] ∈ L for arbitrary tangent vectors g' ∈ TgG.
If you work through the definitions and algebra, you'll also see that [g0g'g1,e'] = Ad(g0)[g',Ad(g1)e'] holds. Thus, it also follows that [Ad(h)g',e'] = [hg'h-1,e'] = Ad(h)[g,Ad(h-1)e] = Ad(h)[g',h-1eh], so that [Ad(h)g',Ad(h)e'] = Ad(h)[g',h-1(he'h-1)h] = Ad(h)[g',e'].
As is the case for all manifolds, associated with G are the cotangent spaces Tg* G and, with this, a natural bi-linear product u ∈ Tg* G, g' ∈ TgG ↦ <u,g'> ∈ ℝ. Correspondingly, from the tangent spaces are inherited a natural product over cotangent spaces on both the left and right u ∈ Tg* G; g0, g1 ∈ G ↦ g0ug1 ∈ Tg0gg1* G such that <g0ug1,g0g'g1> = <u,g'> for g' ∈ TgG.
The dual Lie algebra L* = Te* G is just the cotangent space at the group identity e.
Likewise, the coadjoint action of G on L* is defined for g ∈ G by u ∈ L* = Te* G ↦ gug-1 ∈ Tgeg-1* G = Te* G = L* . Similarly, just as you can define Ad(g)h' = gh'g-1 over tangent spaces, you can also define the coadjoint action CoAd(g)u = gug-1 over all cotangent spaces; and this reduces to a closed operation on L* , as a special case. The coadjoint orbit of u is Ω = { gug-1: g ∈ G}.
Repeated derivatives can be treated the same. The orbit Ω, by virtue of having a product by G on the left and right, passes down a product to its (co-)tangent spaces. By virtue of this, you can then write ω(g-1ug)(X',Y') = <g-1ug,[X,Y]> = <u,g[X,Y]g-1> = <u,Ad(g)[X,Y]> = <u,[Ad(g)X,Ad(g)Y]> = <u,[gXg-1,gYg-1]> = ω(u)(gX'g-1,gY'g-1).
Now, I don't know why mathematicians play coy and obfuscate their formalism by going out of their way to avoid writing things in a natural, obvious and transparent way as I've just done. But I think a large part of it is that what we call "job security". Like the Egyptian scribes who kept a monopoly on orthography by intentionally using an overwrought system even when much better (and more easily teachable) alternatives were eventually available (the Sinai script and Phoenician), they may do it to over-intellectualize the subject and put an intellectual firewall around the subject so as to erect a barrier to entry.
The question you asked would never had to have been asked, if they had just written it down the right way to start, and not put up the obfuscation firewall.
This is, of course, just the tip of the iceberg. Most other fields of mathematics have a firewall of obfuscation erected around them, similar to what you've just seen here. This problem is deeply ingrained in human nature and is, needless to say, why I think the time has come to simply automate the field and remove the human from the picture.
Just like when Logic Theorist ( https://history-computer.com/Library/Logic%20Theorist%20memorandum.pdf ) found the dramatically simpler proof of Pons Asinorum (which I also found and thought was the obvious proof); the kind of simplification you just saw here is what you will see arise as a result. Or perhaps a framework for automation is already operational at this location, and I'm speaking in the past tense with this very description being a product of that automation. :) [Higher Order Logic Theorem Proving https://www.google.com/search?q=Theorem+Provers+in+Higher+Order+Logic ; Fully Automated Theorem Provers https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20050184118.pdf ; Automated Theorem Discovery https://www.google.com/search?q=Automated+theorem+discovery ]