I think you started with some wrong definitions.
The right definition for stabilizability and detectability should be:
A system is stabilizable if the uncontrollable states are stable.
A system is detectable if the unobservable states are asymptotically stable.
You seemed to be using the definition the other way around.
So with these definitions, the task should be very straightforward.
For stabilizability, you simply decompose the system into controllable and uncontrollable subsystems, and then check if the uncontrollable subsystem is stable.
For detectability, again, decompose the system into observable and unobservable subsystems, then check if the unobservable subsystem is asymptotically stable.
Let $x\in \mathbb{R}^n$ and $U$ be an open neighbourhood of $x.$ Let $\Delta$ be a smooth $k$-dimensional distribution on $\mathbb{R}^n.$ When $\Delta$ is involutive, we have, by Frobenius' theorem, that it is locally completely integrable. That means that there exists a coordinate transformation where the immersed submanifolds tangent to $\Delta$ are "flattened" in the new coordinates.
Let us use that change of coordinates. Let the coordinate transformation be denoted $\Phi: U\to V.$ Define our new set of coordinates $$\begin{pmatrix}z_1(x) \\ \vdots \\z_n(x)\end{pmatrix} = z(x) = \Phi(x).$$ The sets tangent to $\Delta$ are immersed submanifolds given in the new coordinates $z$ by fixing the $n-k$ functions $z_{k+1}(x), \dots, z_n(x)$ to any constant. These are your $\lambda$ functions.
It helps to move to these new coordinates $z$ where $\Delta$ is flattened. Notice that the $\Phi$-related distribution of $\Delta$ is generated by the vector fields
$$\partial_{z_1},\dots,\partial_{z_k}.$$
Let us call this distribution (defined on the open set $V$) $\bar{\Delta}.$ Let us also denote the $\Phi$-related vector field of $f$ as $\bar{f}.$
All of this discussion has ignored the important property connecting $f$ and $\Delta.$ Now let us talk about that.
Since $\Delta$ is involutive we also have $\bar{\Delta}$ is involutive. Further, if $[f, \Delta] \subseteq \Delta$ we have that $[\bar{f}, \bar{\Delta}] \subseteq \bar{\Delta}$
Recognize that since $\bar{\Delta}$ is generated by the constant, standard vector fields $\partial_{z_1}, \dots, \partial_{z_k}$ we can say
$$
\begin{aligned}\\
[\bar{f}, \partial_{z_1}] &= \sum_{\ell=1}^{k} c_{1,\ell} \partial_{z_\ell}\\
&~\vdots\\
[\bar{f}, \partial_{z_k}] &= \sum_{\ell=1}^{k} c_{k,\ell} \partial_{z_\ell}
\end{aligned}$$
where $c_{i,j}$ are smooth functions on $V.$ At this point if you write $\bar{f}$ as a smooth function combination of the constant vector fields $\partial_{z_1},\ldots,\partial_{z_n}$ and combine with the above equation, what can you say about the coefficients that multiply the vector fields $\partial_{z_{k+1}},\dots, \partial_{z_n}$? Direct computation should verify that those coefficients cannot be functions of $z_{1}$ through til $z_k.$
Best Answer
For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by
$$ \vec{v}^\top = \begin{bmatrix}0 & \cdots & 0 & 1\end{bmatrix} \begin{bmatrix} B & B\, A & B\, A^2 & \cdots & B\, A^{n-1} \end{bmatrix}^{-1}, \tag{1} $$
$$ T = \begin{bmatrix} \vec{v}^\top \\ \vec{v}^\top A \\ \vec{v}^\top A^2 \\ \vdots \\ \vec{v}^\top A^{n-1} \end{bmatrix}. \tag{2} $$
So using the transformation $z(t) = T\,x(t)$ gives
$$ \dot{z} = T\,A\,T^{-1} z(t) + T\,B\,u(t) + T\,\phi(t) + T\,D(t), \tag{3} $$
where $(T\,A\,T^{-1}, T\,B)$ will be in the controllable canonical form.
If you would like to know more about how to derive this then you can look at this related question.