Let me reply taking M to be the identity (indeed M is somewhat cosmetic to the discussion).
The identity $F\circ Ψ=DΨ\circ F$ is what one considers for example in the Grobman-Hartman theorem, passing from a dynamics to its linearization say at a fixed point. The possibilities for F are endless; $F$ would then be a topological conjugacy, perhaps locally, although maybe not very regular, more precisely at most Hölder continuous in general. Moreover, $F$ need not satisfy any invariance properties, which if I understand correctly is your main concern.
Unfortunately you'll have to face a «no» answer, as there is no general/generic way to handle your problem…
First of all, your definition of «non-hyperbolic» is not clear to me when $n>1$. As far as I know a singular point like yours is deemed hyperbolic when $0$ does not belong to the real convex hull of the spectrum (in $\mathbb C$) of the linear part $A$. In case the singularity is hyperbolic, Poincaré's linearization theorem ensures that the system is dynamically equivalent to its linear part, as you refer to in the question. Certainly when you have null eigenvalues the system is not hyperbolic in that sense.
Now if the singular point is not hyperbolic, then a lot of bad behaviors can crop up, even in the case $n=2$. The presence of resonances ($\mathbb Q$-linear relations between the eigenvalues of $A$) or quasi-resonances (irrational relations which are «well approximated» by rational ones) can mess things up. Even from a formal viewpoint the answer is not clear, and certainly not computable without restrictive conditions on $f$ (e.g. polynomial, and even then…). In particular it is not true that the system $\dot x=f(x)$ is conjugate (dynamically equivalent) to a system where $f$ is algebraic using changes of coordinates which are merely $C^1$ at the singular point. Hence it is not sufficient to take a finite jet (let alone the second jet) to answer your question when there are resonances (as is the case here when at least one eigenvalue vanishes).
Dynamical systems with at least one zero eigenvalue are called «saddle-nodes», so you might wish to search for this keyword. Notice, though, that the theory of saddle-nodes is complete only in the smooth case (that is, $f$ smooth) for $n=2$, and partially covered when $n>2$. The more resonances (e.g. zero eigenvalues) the more difficult the problem is, even from a purely theoretical viewpoint…
So, without more information on your specific system, nothing can be said.
EDIT: You still can say something regarding stability. For instance look at the system $$\dot x(t) = x(t)^{k+1} \\ \dot y(t)=-y(t)$$ with $x, y\in\mathbb R$. Depending on the parity of $k$, either all trajectories are stable or only a single trafectory is (that's $y=0$). In these problems you need to identify directions in the $x$-variable where you have contributions of exponential terms $\exp(x^{-k}/k)$. This can be done by inspecting the linear part, and in general governs «how stable» the solutions will be. I don't have a recipe when $n>1$, though.
Best Answer
Relaxing the hypothesis that $f$ is invertible, a sufficient condition to linearize globally the system with a change of variables $f$ is that $F$ is analytic. Then the Carleman linearization technique and other similar techniques (as can be seen in the book "Nonlinear dynamical systems and Carleman Linearization") can be applied to transform the nonlinear system in a infinite dimensional linear system.