I assume that you consider the infinite cyclic group to be free. Then take the free nilpotent group $G_n$ of class $c\gg 1$ with 2 generators. It has an infinite cyclic central subgroup $K_n$, the factor-group $G_{n-1}=G_n/K_n$ is nilpotent and torsion-free, it has an infinite cyclic subgroup $K_{n-1}$, and so on. Every group $G_i$ is 2-generated, the chain can be arbitrary long ($n$ depends on $c$).

Edit 1: Since you do not want to consider $\mathbb Z$ free enough, here is another example. Take the Baumslag-Solitar group $BS(2,3)=\langle a,t | ta^2t^{-1}=a^3\rangle$. It is non-Hopfian, and has a free normal subgroup $K$ such that $BS(2,3)/K$ is isomorphic to $BS(2,3)$. So all $G_i$ are isomorphic to $BS(2,3)$, all $K_i$ are infinitely generated free groups. Is that what you want?

Edit 2: Since you have another condition now, "commutative transitive", then here is another example. Take a non-elementary torsion-free hyperbolic group $G$. It has a free normal subgroup $N$ such that $G/N$ is still non-elementary torsion-free and hyperbolic (that is proved by Olshanskii, and also by several others, including Delzant). You can continue as long as you wish. All $G_i$ will be hyperbolic and torsion-free (hence commutative-transitive), all $K_i$ will be infinitely generated free groups.

Just to anticipate a future change in the formulation of the question: if you really insist that $K_i$ are finitely generated, the question becomes harder and I am not sure the answer is still the same.

Edit 3: Since you want to have an infinite sequence, here is what to do. For every hyperbolic group $G$ with 2 generators, there exists another 2-generated hyperbolic group $G'$ and a free normal subgroup $N \le G'$ such that $G'/N=G$. This can be done using Rips' construction. In the original Rips' construction (and in all modifications) $N$ was not free, because he wanted $N$ to be finitely generated. But if you do not want $N$ to be finitely generated, it is easy to modify Rips' construction to make $N$ free. Using this you can construct your sequence $G_0=G, G_1=G', G_2=G_1', ...$.

Edit 4:In fact Rips' construction does not quite work because the number of generators increases. Certainly if $G$ is free of rank 2, $G'$ cannot be of rank 2. But here is another construction. Take a (torsion-free) lacunary hyperbolic group given by an infinite presentation satisfying a small cancelation condition (see http://front.math.ucdavis.edu/0701.5365). Let $r_1,r_2,...$ be the presentation of $G$. Then $G$ is commutative transitive (it is easily deduced from the fact that $G$ is an inductive limit of hyperbolic groups and surjective homomorphisms). Now the group $G'$ given by the same presentation but without $r_1$ is again lacunary hyperbolic, $G$ is a factor-group of $G'$ over the normal subgroup $N$ generated by $r_1$. It is possible to prove that $N$ is free. Indeed, if some product of conjugates of $r_1$ is equal to 1 in $G$, consider the corresponding van Kampen diagram. The boundary of that diagram has parts labeled by $r_1$ and parts labeled by the conjugators. By Greendlinger lemma, if the diagram has cells, it must have a cell with more than, say, $90\%$ of its boundary common with the boundary of the diagram (take the small cancelation condition $C'(1/300)$). Then more than a half of that part of the boundary must be inside a conjugator, the conjugator can be shortened, and a shorter product of conjugates of $r_1$ is equal to 1 in $G'$. Since $G'$ is lacunary hyperbolic again and satisfies the same small cancelation condition as $G$, we can repeat the construction. Since the presentation is infinite, the process will continue indefinitely.

Edit 5: A more clean way to prove that$N$ is free in Edit 4 is the following. suppose that some product of conjugates of $r_1$ is equal to 1 in $G'$. Consider the corresponding van Kampen diagram $\Delta$. Its boundary label is equal to 1 in the group given by 1 relator $r_1$. Consider the diagram $\Psi$ corresponding to that equality. Now identify the boundaries of $\Psi$ and $\Delta$. We get a diagram over the presentation of $G$ on a sphere: $\Delta$ occupies the northern hemisphere, $\Psi$ occupies the southern hemisphere, and the product of conjugates of $r_1$ labels the equator. Reduce that diagram. Since we can assume that $\Psi$ is reduced, and the $r_1$-cells are only in the south, the $r_1$-cells won't cancel. Hence we shall have a reduced non-empty diagram over the presentation of $G$ on a sphere which is impossible because of the Greendlinger lemma (the boundary of the spherical diagram is empty).