When you have an observable, the eigenvectors with different eigenvalues are orthogonal, but your second set does not require that the vectors be orthogonal. And since the elements of the second set are not required to be normalized we learn very very little (almost nothing) about how $A$ scales any particular vector, or even whether it scales the vector or does something else.
And all your conjectures are false and they all have counterexamples in two dimensions (they are all true in one dimension).
Consider $A=\sigma_z=\left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $ \Psi_2=\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]$ but we can consider a $\Phi_1=\left[\begin{matrix} a \\ b \end{matrix}\right]$ and all we need is that $a^*a-b^*b=1.$ This is a counter example to your generalization for $M<N$ since you could have for instance $\Phi_1=\left[\begin{matrix} \sqrt 2 \\ 1 \end{matrix}\right].$
But this also has zero relationship with the cited problem. Your nongeneralized problem is also false and again for reasons related to lack of orthonormalization of $\Phi_k$ and that also have no bearing on the problem cited. For example consider $A=\left[\begin{matrix} 4 & 0 \\ 0 & 9 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $ \Psi_2=\left[\begin{matrix} 0 \\ 1 \end{matrix}\right]$ but we can consider $\Phi_1=\left[\begin{matrix} 0 \\ a\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} b \\ 0 \end{matrix}\right]$ so now the requirements are that $9a^*a=4$ and $4b^*b=9$ so we can for instance use $\Phi_1=\left[\begin{matrix} 0 \\ 2/3\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} 3/2 \\ 0 \end{matrix}\right]$ and note that this is a counter example to your question but satisfies the cited question. An even simpler counter example would be just to multiply $ \Psi_1$ by $-1.$
Which means your question is wrong, but the cited question is still up in the air. The cited question makes no claim about nondegenerate spectra or about how many vectors you have. And it isn't even clear that the spectra has to be real because it just days operator, not observable. But we can find a counterexample where it is an observable and has a nondegenerate spectra.
If all you want to do is understand the notation think of kets as column vectors, operators as square matrices and bras as row vectors, then think of dagger as taking the transpose of the matrix and then the complex conjugate of evey element of the matrix. Think of the bra version of a ket as the dagger of the ket. That's what you need to see what is going on.
But back to your cited question. If you have a bunch of orthonormal $\Psi_k$ that are eigenvectors of $A$ then this doesn't tell us anything at all about what happens outside the (closure) of the linear span of those eigenvectors, and we can definitely find a counter example if the $\Psi_k$ live in some proper subspace of the Hilbert space. So either that was a counterexample we were supposed to find, or else let's assume we have lots and lots of eigenvectors, $\Psi_k,$ say enough to span a dense subset of the Hilbert space. We can then actually write $A=B+iC$ for some observables $B$ and $C$ ($C=0$ if A is an observable). And we can then do it so $B$ and $C$ have common eigenvectors with the real and imaginary parts of the $a_k$ corresponding to the eigenvalues of $\Psi_k$ with $B$ and $C$ respectively.
OK now we want to consider whether the $\Phi_k$ are eigenvectors of $A.$
This again has a counterexample in the land of 2x2 matrices. Let $A=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ with $\Psi_1=\left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$ and $ \Psi_2=\left[\begin{matrix} 0\\ 1 \end{matrix}\right]$ but we can consider $\Phi_1=\left[\begin{matrix} 1 \\ 1\end{matrix}\right]$ and $\Phi_2=\left[\begin{matrix} 0\\ 1 \end{matrix}\right]$ it satisfies all the conditions but $\Phi_1$ is not an eigenvector of $A.$
Each time I just thought square matrix when I saw an operator, thought column vector when I saw a ket, thought row vector when I saw a bra, and thought transpose and conjugate when I saw a dagger and thought dagger of when a saw a bra version of a ket.
There can be some differences but they really come down to just extra details about infinite dimensional spaces and that is just because of unfamiliarity with infinite dimensional spaces not because they are weird or different the nice infinite dimensional spaces are nice and the finite dimensional spaces just happen to already be nice so you aren't used to worry about whether something is nice.