= Solution
Use units $c=1$ and take the initially expanding branch with ordinary positive matter or radiation density. In the <Friedmann-Lemaître-Robertson-Walker metric>, $k=+1,0,-1$ respectively describe positive, zero and negative spatial curvature. Separately conserved <pressureless matter> has $\rho_M\propto a^{-3}$ and <radiation in cosmology> has $\rho_R\propto a^{-4}$, so either dominates the curvature and vacuum terms sufficiently far back on a hot Big Bang branch.
With negligible <cosmological constant>, a spatially flat radiation universe has $a\propto t^{1/2}$ and a flat matter universe has $a\propto t^{2/3}$; both expand forever while decelerating. An open universe also expands forever and eventually becomes approximately curvature dominated, with $a\propto t$. A closed universe containing only ordinary matter/radiation reaches a maximum size and recollapses. For example, dust gives $\dot a^2=C/a-k$ with $C>0$, so only $k=+1$ admits the positive turning point $a=C$.
\b[Curvature alone does not determine the fate when the cosmological constant is retained.] Positive $\Lambda$ can cause late accelerated expansion and prevent even a closed universe from recollapsing; whether a closed expanding branch turns around depends on the relative matter and vacuum terms. Negative $\Lambda$ can cause recollapse even for flat or open spatial sections. These statements follow by inspecting $\dot a^2=C/a-k+\Lambda a^2/3$ for dust; the curvature sign is a geometrical classification, not by itself a universal destiny criterion.
Define the <critical density> $\rho_c=3H^2/(8\pi G)$ and <cosmological density parameters> $\Omega_M=\rho_M/\rho_c$, $\Omega_R=\rho_R/\rho_c$, $\Omega_\Lambda=\Lambda/(3H^2)$ and $\Omega_k=-k/(a^2H^2)$. The <Friedmann equation> gives $\Omega_M+\Omega_R+\Omega_\Lambda+\Omega_k=1$. Dividing the <Friedmann acceleration equation> by $-H^2$ gives
$$
q=\frac{4\pi G}{3H^2}(\rho+3P)-\Omega_\Lambda
=\frac12\Omega_M+\Omega_R-\Omega_\Lambda.
$$
Consequently
$$
\boxed{q=\frac12\Omega_M-\Omega_\Lambda\quad\text{for dust plus vacuum},\qquad q=\Omega_R-\Omega_\Lambda\quad\text{for radiation plus vacuum}.}
$$
The <deceleration parameter> measures the sign and size of the acceleration relative to $aH^2$: $q>0$ means deceleration, $q<0$ accelerated expansion and $q=0$ instantaneously zero acceleration. In particular a falling <Hubble parameter> alone does not imply deceleration, since $q=-1-\dot H/H^2$.
For conformal time $d\eta=dt/a$, the <conformal Hubble parameter> is $\mathcal H=aH$. Using matter conservation and differentiating $\Omega_M\propto\rho_M/H^2$,
$$
\frac{\dot\Omega_M}{\Omega_M}=-3H-2\frac{\dot H}{H}=H(2q-1),\qquad
\boxed{\Omega_M'=\mathcal H\Omega_M(\Omega_M-1-2\Omega_\Lambda).}
$$
This is the <matter density flow with a cosmological constant>. Neglecting $\Omega_\Lambda$ during matter domination reduces it to the requested relation
$$
\boxed{\Omega_M'=\mathcal H\Omega_M(\Omega_M-1).}
$$
That last equation is not exact with nonzero $\Lambda$. For example, a flat dust-plus-vacuum state with $(\Omega_M,\Omega_\Lambda)=(0.9,0.1)$ gives $\Omega_M'/\mathcal H=-0.27$, whereas the matter-only expression gives $-0.09$.
For the <flatness problem>, let $\Omega_{\rm tot}=\Omega_M+\Omega_R+\Omega_\Lambda$. Curvature satisfies $\Omega_{\rm tot}-1=k/(aH)^2$, so
$$
\frac{d\log|\Omega_{\rm tot}-1|}{d\log a}=2q.
$$
In decelerating near-flat radiation and matter eras, deviations therefore grow as $a^2$ and $a$ respectively. Maintaining small curvature at a late epoch requires very small early deviations unless a prior mechanism suppresses them. <Cosmic inflation>, with $q\simeq-1$, drives the deviation down approximately as $a^{-2}$. If vacuum energy is present, flatness means $\Omega_{\rm tot}=1$, not $\Omega_M=1$.
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