One can understand the image of $\psi$ or $\psi|_M$ in terms of the inverse map $\phi$.
$\phi$ maps an $m$ set to an $(m+1)$ set (if possible) in the following way. If $$A=\{i_1,\dots,i_m\}$$ is our $m$-set, we again look at the set of indices for which $i_j -2j$ is minimized $$N(A)=\{j\in\{0,1,\dots,m\}\mid i_j -2j=\min\},\quad i_0:=0$$ but this time we take the maximum of all indices and add 1 on the value to get our $(m+1)$-set:$$m(A)=\max N(A)\\ \phi(A)=\{i_1,\dots,i_{m(A)},i_{m(A)}+1,i_{m(A)+1},\dots,i_{m}\}.$$ So $\phi$ is defined iff $i_{m(A)}\neq n$. Moreover: $$\mathrm{Im}(\psi)=\mathrm{Def}(\phi)$$
So now it is easy to generalize this for $\psi|_M$: A subset $M'=\{i_1,\dots,i_l\}$ of $M$ is not in the image of $\psi|_M$ iff $i_{m(M')}+1\notin M$.