Actually, you have it backwards. It's more intuitive and less arcane and (as Kirillov himself noted) it may have been how Lie originally tried to frame his formalism. This is best seen with an example.

$\newcommand\so{so}\DeclareMathOperator\SO{SO}$The Lie algebra $\so(3)$ for the group $\SO(3)$ may be presented as being generated by the basis $(X, Y, Z)$, subject to the identities
$$[Y, Z] = X, \quad [Z, X] = Y, \quad [X, Y] = Z.$$
In the dual $\so(3)^*$ the role of basis elements and coordinates is reversed, so that $(X, Y, Z)$ are now thought of as coordinates (with respect to the dual basis). The Lie bracket becomes a Poisson bracket
$$\{Y, Z\} = X, \hspace 1em \{Z, X\} = Y, \hspace 1em \{X, Y\} = Z,$$
over the function space $C^\infty\left(so(3)^*\right)$, which contains the linear functions, i.e. $\so(3)^{**}$, which is equivalent to $\so(3)$, itself, so that the Poisson bracket is an *extension* of the Lie bracket to non-linear functions.

Associated with the Lie group $\SO(3)$ is a family of symmetry transformations generated by infinitesimal transformations that correspond to the elements of $\so(3)$. Over this Poisson manifold, each $\Lambda = xX + yY + zZ$ gives rise to an infinitesimal transformation $\Delta = \{\cdots, \Lambda\}$, whose actions on the basis/coordinate-functions is:
$$\Delta X = yZ - zY, \hspace 1em \Delta Y = xZ - zX, \hspace 1em \Delta Z = xY - yX.$$
In the language of mathematicians, $\Delta$ is a vector field and its exponential $e^{s\Delta}$, its flow, is the action of the identity component of the Lie group (here: the Lie group itself, since it's connected) on the space. Here, it can be tabulated as follows:
$$\Delta^2(X, Y, Z) = (x, y, z) (xX + yY + zZ) + \lambda (X, Y, Z),$$
where $\lambda = -\left(x^2 + y^2 + z^2\right)$,
$$\Delta (xX + yY + zZ) = 0, \quad \Delta^3(X, Y, Z) = \lambda\Delta(X, Y, Z),$$
so
$$\begin{align}
e^{s\Delta}(X, Y, Z) &{}= C(X, Y, Z) \\
&{}+ S(yZ - yZ, xZ - zX, xY - yX) \\
&{}+ D(x, y, z)(xX + yY + zZ) + D\lambda (X, Y, Z),
\end{align}$$
where the functions $C(\lambda, s)$, $S(\lambda, s)$ and $D(\lambda, s)$ are given by
$$(C, S, D)_{s=0} = (1, 0, 0),\hspace1 em\Delta(C, S, D) = \frac{\partial}{\partial s}(C, S, D) = (\lambda S, C, S),$$
which, for $\lambda < 0$, work out to:
$$(C, S, D) = \left(
\cos\left(\sqrt{-\lambda}s\right), \frac{\sin\left(\sqrt{-\lambda}s\right)}{\sqrt{-\lambda}},
\frac{1 - \cos\left(\sqrt{-\lambda}s\right)}{\lambda}\right).$$

Since this action is actually taking place over the function space of $\so(3)^*$, what is denoted by the coordinates $(X, Y, Z)$, in these expressions, are actually linear functions, i.e. elements of $\so(3)^{**}$ — effectively the basis elements of $\so(3)$, itself. The result is that the adjoint action of $\so(3)$ by the group $\SO(3)$ is extended to non-linear functions.

This allows you, for instance, to talk about *invariants* in an *intrinsic* way; i.e. without recourse to matrix or other representations. For example, you have:
$$X^2 + Y^2 + Z^2$$
which you can now *directly* show is an invariant, since, under transformation:
$$\begin{align}
\Delta\left(X^2 + Y^2 + Z^2\right) &{}= 2X\Delta X + 2Y\Delta Y + 2Z\Delta Z \\
&{}= 2(X(yZ - zY) + Y(zX - xZ) + Z(xY - yZ)) \\
&{}= 0.
\end{align}$$

The locus of all transforms $\Delta = \{\cdots, \Lambda\}$ corresponding to all $\Lambda$ that $\so(3)$ gives rise to is one and the same as the symplectic leaf generated from the point $(X, Y, Z)$ that these transforms act on. Here: we see that it's a sphere with radius $X^2 + Y^2 + Z^2 > 0$ if $(X, Y, Z) \ne 0$ and the fixed point $(0, 0, 0)$ if $(X, Y, Z) = (0, 0, 0)$.

All of this involves non-linear functions. You can't directly address any of it with the Lie algebra $\so(3)$, since that's a linear space.

So, the coadjoint orbits — and the related Poisson manifold machinery — together provide a way to talk about the Lie algebra $\so(3)$ as a *non-linear* space. As Kirillov noted, it appears that Lie was originally striving for a non-linear theory. So, it ties off a long-standing loose end.

It also provides a way to by-pass much of the machinery associated with Lie algebra contractions. Consider, for instance, a 3-parameter family of Lie groups whose basis elements comprise the following 11 generators (with 9 of them arranged together as 3-vectors, for convenience):$\newcommand\bJ{\mathbf J}\newcommand\bK{\mathbf K}\newcommand\bP{\mathbf P}\newcommand\bW{\mathbf W}\newcommand\bomega{\boldsymbol\omega}\newcommand\bupsilon{\boldsymbol\upsilon}\newcommand\bepsilon{\boldsymbol\epsilon}$
$$\bJ = \left(J_0, J_1, J_2\right), \quad \bK = \left(K_0, K_1, K_2\right), \quad \bP = \left(P_0, P_1, P_2\right), \quad H, \quad M,$$
whose Lie brackets give rise to a 3-parameter family of Poisson brackets identified by the following transform law:
$$\begin{align}
\Delta\bJ &{}= \bomega\times\bJ + \bupsilon\times\bK + \bepsilon\times\bP, \\
\Delta\bK &{}= \bomega\times\bK - \gamma\bupsilon\times\bJ + \bepsilon M - \beta\tau\bP, \\
\Delta\bP &{}= \bomega\times\bP - \bupsilon M + \lambda\bepsilon\times\bJ - \kappa\tau\bK, \\
\Delta H &{}= -\beta\bupsilon\cdot\bP - \kappa\bepsilon\cdot\bK, \\
\Delta M &{}= -\gamma\bupsilon\cdot\bP - \lambda\bepsilon\cdot\bK
\end{align}$$
corresponding to $\Delta = \{\cdots,\Lambda\}$, with $\Lambda = \bomega\cdot\bJ + \bupsilon\cdot\bK + \bepsilon\cdot\bP - \tau H + \psi(M - \alpha H)$. The parameters are $(\alpha, \beta, \kappa)$, with $(\gamma, \lambda) = (\alpha\beta, \alpha\kappa)$, with its invariants including the fundamental invariants:
$$\begin{align}
\mu &{}= M - \alpha H, \\
\nu &{}= W^2 - \gamma W_0^2 + \lambda W_4^2, \\
\rho &{}= \beta P^2 - 2HM + \alpha H^2 - \kappa K^2 + \alpha\beta\kappa J^2,
\end{align}$$
as well as derived invariants, such as
$$\mu^2 - \alpha\rho = M^2 - \gamma P^2 + \lambda K^2 - \gamma\lambda J^2,$$
where
$$W_0 = \bJ\cdot\bP, \quad \bW = \left(W_1, W_2, W_3\right) = M\bJ + \bP\times\bK, \quad W_4 = \bJ\cdot\bK.$$

They can actually all be combined into a *single* Poisson manifold — by just throwing in $(\alpha, \beta, \kappa)$ as extra coordinates. Together, this comprises the coadjoint orbits of *all* the members of the Bacry Lévy–Leblond classification of possible kinematic groups, with $\gamma > 0$ and $\kappa = 0$ being the Poincaré group, centrally extended with a trivial central extension by $\mu$ and with $\alpha = 0$, $\beta \ne 0$ and $\kappa = 0$ being the Bargmann group, which is the (non-trivial) central extension of the Galilei group.

What was originally Lie algebra contractions, involving the parameters $(\alpha, \beta, \kappa)$, are now just continuous transitions in the coordinates $(\alpha, \beta, \kappa)$ of the combined Poisson manifold.

The formalism is more intuitively grounded, as well. The components of $\Lambda$ in the transform $\Delta = \{\cdots, \Lambda\}$ are:

- infinitesimal rotations $\bomega = (x, y, z)$, which we've already discussed,
- infinitesimal boosts by an infinitesimal change $\bupsilon$ in velocity,
- infinitesimal spatial translations $\bepsilon$,
- infinitesimal time translations $\tau$ (with the negative on $H$ being a legacy convention),
- infinitesimal actions $\psi$ for the central charge $\mu = M - \alpha H$.

The transforms give you the transform laws for the generators themselves under the respective actions, e.g.

$\Delta\bK = \bepsilon M$ under a spatial translation

$\Lambda = \bepsilon\cdot\bP$ (which shows that

$\bK$ is an "

$M$-moment"), or

$\Delta\bP = -\bupsilon M$ under a boost

$\Lambda = \bupsilon\cdot\bK$ (which shows that

$\bP$ is an "

$M$-momentum").

It only seems "arcane" because it's usually being presented only in the abstract; but when get down into the details to see what it's actually all about, without the abstraction, you see that it's all quite intuitive, and intuitively grounded.

sameas the adjoint representation (at least for semisimple groups)! In positive characteristic they can differ, and the coadjoint representations can be more useful; see, for example, Kac and Weisfeiler - Coadjoint …. $\endgroup$