Answer: $2 : 5$.
Let the segment $MD$ intersect the side $AC$ at $N$ and the diagonal $AC$ at $P$.
From $BC \parallel AD$ we get $\angle CBK = \angle BKA$.
Since $BK$ is the bisector, $\angle ABK = \angle CBK$
hence $\angle ABK = \angle BKA$ and the triangle $ABK$ is isosceles with $AB = AK$. From $2AB = AD$ it follows that $K$ is the midpoint of $AD$, similarly $N$ is the midpoint of $BC$. It is clear that $BNDK$ is a parallelogram, therefore $ND = BK$ and $NP$, $LK$ are the middle lines of the triangles $BCL$ and $APD$ respectively. Thus $2NP = BL$ and $2LK = PD$ whence $NP + PD = ND = BK = BL + LK$ and therefore $2NP = 2LK = PD$. Since $N$ is the midpoint of $BC$, $BN = 0.5BC = 0.5AD$. And since $BC \parallel AD$, $BN$ is the middle line of the triangle $AMD$, so $N$ is the midpoint of $MD$, i.e.

$$MN = ND = NP + PD = \frac{3}{2}PD \implies MP = MN + NP = \frac{3}{2}PD + \frac{1}{2}PD = 2PD. \tag{1}$$
Let $Q$ be the intersection point of the side $AD$ and the line passing through $P$ parallel to $MF$. By Thales' theorem $FQ : QD = MP : PD \stackrel{(1)}{=} 2 : 1$. Since $LK$ is the middle line of the triangle $APD$, $AK = KD$. Therefore
$$AF : AD = AF : (AF + FQ + QD) = FQ : (2FQ + QD) = 2QD : 5QD = 2 : 5.$$