Not really.
The adjoint operator can be defined for arbitrary topological vector spaces; adjugate requires finite-dimensioned spaces.
The adjoint operator is, for real matrices, just the transposed matrix. The adjugate, as you see, has an entirely different definition.
As per your question $2$, $adj(T)$ is the adjugate operator.
Sadly, there is no intuition to be learned, only that matrix differential calculus has inconsistent notation.
I wouldn't see $\frac{\partial w}{\partial w}$ as something that "does transposition": the result you mention follows from the linearity of differentiation and from using the denominator convention.
By linearity of differentiation,
$$A \frac{\partial w}{\partial w} = \frac{\partial A w}{\partial w}$$
If we follow numerator convention, this is the Jacobian matrix of the linear function $Aw$, that is a matrix whose $i,j$th element is
$$\frac{\partial A_i^T w}{\partial w_j} ,$$
where $A_i^T$ is the $i$th row of $A$. Then,
$$\left[\frac{\partial A w}{\partial w}\right]_{i,j} =\frac{\partial }{\partial w_j} \sum_{k=1}^m A_{i,k}w_k=A_{i,j},$$
where $A_i^T$ is the $i$th row of $A$, so
$$A \frac{\partial w}{\partial w} = A.$$
If we instead follow denominator convention, then your expression means the gradient of the linear function $Ax$, that is the matrix whose $i,j$ element is
$$\frac{\partial A_j^T w}{\partial w_i}$$
Then
$$\left[\frac{\partial A w}{\partial w}\right]_{i,j} =\frac{\partial }{\partial w_i} \sum_{k=1}^m A_{j,k}w_k = A_{j,i},$$
so
$$A \frac{\partial w}{\partial w} = A^T.$$
The fact that the author uses a transposed vector at the denominator seems to indicate the numerator convention, however the transposed result hints at a denominator convention. Confusing indeed!
Best Answer
In the context of complex vector spaces, they are different: the adjoint matrix is the conjugate of the transpose matrix.