Intuitive explanation of $Ad,ad,Ad^*,ad*$. Let $G$ be a Lie group. Then the mapping $\Psi_g:G \rightarrow G$ given by $\Psi_{g}(h)=g h g^{-1}$ is a Lie group homomorphism, which I think of intuitively as measuring the "non-commutativeness" of $g$. Taking the differential of $g$ at $e\in G$ we get the Adjoint representation $\mathrm{Ad}_{g}=\left(d \Psi_{g}\right)_{e}: T_{e} G \rightarrow T_{e} G$ which gives an infinitesimal version of the "non-commutativeness" of $g$. The adjoint representation I think of as the (vector field)Lie bracket of left-invariant fields has a good interpretation as the infinitesimal "non-commutativeness" of the flows of left invariant vector fields.
When it comes to dualising the operators to try and get a handle on $Ad^*$ and $ad^*$ my (already) shaky intuition starts to leave me. Is there a way to describe, in words, what these representions are? 
 A: I'd like to share my intuition for the adjoint and coadjoint actions: by specializing to the case of $G = \mathrm{SO}(3)$, where I can visualize things concretely in terms of rotations of $\mathbb{R}^3$. 
Recall that in this case, the Lie algebra $\mathfrak{so}(3)$ of $G$ is given by the $3\times 3$ real skew-symmetric matrices $A$, that is, those matrices satisfying $A^T = -A$. We can write these all out explicitly as:
\begin{equation}
A = \begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ a_2 & a_1 & 0
\end{pmatrix}
\end{equation}
and hence identify $A \in \mathfrak{so}(3)$ with the vector $(a_1, a_2, a_3) \in \mathbb{R}^3$. Then you should check that under this identification, the adjoint action of $\mathrm{SO}(3)$ on $\mathfrak{so}(3)$ becomes the usual action of $\mathrm{SO}(3)$ on $\mathbb{R}^3$ by rotation!
Ok, so what about the duals of these representations? This is where we need the distinction between active and passive transformations: when I consider an element of $\mathrm{SO}(3)$ acting on $\mathbb{R}^3$, I can either imagine it rotating a vector with respect to a fixed orthogonal coordinate system (the active perspective), or rotating an orthogonal coordinate system around a fixed vector (the passive perspective). In fact, these two perspectives are dual to each other in a precise sense: once we have chosen a basis for $\mathfrak{so}(3)$ as above, the coadjoint action of $\mathrm{SO}(3)$ on $\mathfrak{so}(3)^{\ast}$ is exactly given by the passive rotations of $\mathbb{R}^3$ (while the adjoint representation is given by the active transformations). You can check this by observing that if a vector $v$ has coordinates $(a_1, a_2, a_3)$ in a given orthogonal basis $e_1, e_2, e_3$ of $\mathbb{R}^3$, then $v$ has coordinates $T^{-1} (a_1, a_2, a_3)$ with respect to the orthogonal basis $T e_1, T e_2, T e_3$, for any $T \in \mathrm{SO}(3)$.
This intuition generalizes to any Lie group $G$, which you can think of as acting on a generalized rigid body in some sense. I hope this helps!
