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.
The differential of a function $\mathrm{d}f$ is its best linear approximation at some point $x_0$. In this case $\det$ is a function $\det:M_{n\times n} \rightarrow\mathbb{R}$ where $M_{n\times n}$ is the space of $n\times n$ square matrices. Therefore, a matrix is the equivalent of a point for real functions. The best linear approximation to $\det$ near the identity is given by:
$$ \det(\mathbf{I}+\mathbf{M})=\det(\mathbf{I})+\mathrm{d}(\det(\mathbf{I}))\mathbf{M}+R(\mathbf{I},\mathbf{M}),\qquad \lim_{\|\mathbf{M}\|\rightarrow 0} \frac{R(\mathbf{I},\mathbf{M})}{\|\mathbf{M}\|}= 0 \tag{1}\label{eq1}.$$
The analogous expression for $f: \mathbb{R} \rightarrow\mathbb{R}$ is
$$ f(x_0+\epsilon)=f(x_0)+\mathrm{d}(f(x_0))\epsilon+R(x_0,\epsilon),\qquad \lim_{|\epsilon|\rightarrow 0} \frac{R(x_0,\epsilon)}{|\epsilon|}= 0 \tag{2}\label{eq2}.$$
Regarding the original question, $\det'(\mathrm{I})=\mathrm{tr}$ is equivalent to the following:
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \mathrm{tr}(\mathbf{B}) \tag{3} \label{eq3}$$
You can use induction to prove $\eqref{eq3}$. The first step is to prove it for $n=2$:
$$\frac{\mathrm{d}}{\mathrm{d}t} \left . \begin{bmatrix}
1+tB_{11} & tB_{12} \\
tB_{21} & 1+tB_{22}
\end{bmatrix} \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left[
(1+tB_{11})(1+tB_{22}) - t^2B_{12}B_{21} \right] \right|_{t=0}=B_{11}+B_{22}. \tag{4} \label{eq4}$$
Assuming that $\eqref{eq3}$ is valid for $n=k-1$, we will prove it for $n=k$.
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [
\sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t)
\right] \right|_{t=0} \tag{5} \label{eq5}$$
here $M_{i,k}(t)$ represent the $(i,k)$ minor of $\mathbf{I}+t\mathbf{B}$.
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [
\sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t)
\right] \right|_{t=0}=\sum_{i=1}^k\left. \left [
B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t)
\right] \right|_{t=0} \tag{6} \label{eq6}$$
evaluating the first term in the RHS of $\eqref{eq6}$ at $t=0$ gives zero except when $i=k$ since each minor has lower order proportional to $t$. On the other hand, $M_{k,k}(0)=\det(\mathbf{I})=1$. Therefore,
$$ \sum_{i=1}^k\left. \left [
B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t)
\right] \right|_{t=0}=B_{k,k}+\frac{\mathrm{d}}{\mathrm{d}t} M_{k,k}(0)= \sum_{j=1}^k B_{j,j}\tag{7} \label{eq7}$$
In the last step we use the hypothesis $n=k-1$ since $M_{k,k}$ is $\det(\mathbf{I^{(k-1)\times(k-1)}}+t\mathbf{B^{(k-1)\times(k-1)}})$.
Best Answer
When interpreting the exponential for a linear map $f:V\to V$ on a finite dimensional real vector space $V$, it seems better to me not to look at $e^f$ but at the family $e^{tf}$ for $t\in\mathbb R$. This is the solution operator for the linear first order ODE $x'=f(x)$, i.e. the unique solution with initial condition $x(0)=x_0$ is given by $x(t)=e^{tf}(x_0)$. Hence $e^f$ can be interpreted as computing the dependence of the solution at time $t=1$ on the initial condition. At this stage you can use the interpretation of the determinant via volumes you suggested.
Now the properties of the exponential readily imply that $t\mapsto \det(e^{tf})=:\phi(t)$ satisfies $\phi(t+s)=\phi(t)\phi(s)$ and $\phi(0)=1$, which easily implies that $\phi'(t)=\lambda\phi(t)$, where $\lambda=\phi'(0)$. The unique solution of this ODE with initial condition $\phi(0)=1$ is $\phi(t)=e^{\lambda t}$. Hence we conclude that $\det(e^{tf})=e^{t\lambda(f)}$ for some function $\lambda:L(V,V)\to\mathbb R$.
It is also not difficult to see that $\lambda$ has to be invariant under conjugation: Observe that $c(t)=e^{tf}$ is uniquely determined by $c(0)=id$ and $c'(t)=f\circ c(t)$. Now if $g:V\to V$ is a linear isomorphism, it is easy to conclude that $e^{t(g\circ f\circ g^{-1})}=g\circ e^{tf}\circ g^{-1}$. Taking determinants, you conclude that $\lambda(g\circ f\circ g^{-1})=\lambda(f)$. With a bit more effort, you can show that $\lambda$ is linear, which implies that it is a multiple of the trace. Checking the case $f=id$ then sorts out that $\lambda(f)=tr(f)$. Unfortunately, I am not aware of an "intuitive" interpretation of the trace in this setting, apart that it can be viewed as computing the unique $GL(V)$-invariant projection of $f$ onto multiples of the identity.