You can understand in a simple way the factor $1/3$ which gives you the approximate solution in the low frequency regime (more on this later) in the following way. Start by writing the kinetic energy of your system as:
$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \int_0^\ell \frac{1}{2} \lambda \dot{\delta}(u)^2 du$$
where $\delta(u)$ is the displacement of the point of the spring which is in the $x=u$ position in the equilibrium configuration and $\lambda=m_0/\ell$ the linear mass density of the spring. The spring has length $\ell$ when unstretched.
If you suppose an harmonic motion for the mass at a very low frequency, the stretching of the spring will be approximately uniform, which means
$$\delta(u) = \frac{u}{\ell} \delta(\ell)$$
Accepting this approximation by substituting in the expression for kinetic energy one gets
$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \frac{1}{2} \frac{m_0}{\ell} \dot{\delta}(\ell)^2 \int_0^\ell \frac{u^2}{\ell^2} du$$
and after an integration
$$K = \frac{1}{2} \left( m +\frac{1}{3} m_0 \right) \dot{\delta}(\ell)^2$$
which is the expected result.
The system has an infinite number of degrees of freedom, which means that it will have an infinite number of oscillation modes. If $m\gg m_0$ the lowest frequency mode will be approximately described as an oscillation of the mass with an uniform stretch of the spring. In the higher frequency modes the mass will be nearly fixed, and there will be a nearly stationary elastic wave on the spring.
The flaw in your reasoning consist in supposing that the external force applied to the system mass+spring is $-k(x-x_0)$. The applied force is really the tension of the spring at his fixed point, which is not $-k(x-x_0)$ for a spring with mass when there are accelerations.
There is tension in the spring. It it extended and hence there is tension! It is the centre of mass that falls with acceleration $g$ rather then each individual mass. So the equation $$mg-T=mg$$ is invalid. As the two masses fall they will oscillate (getting closer and further away) and the tension will cycle.
Let us call the distance fallen by mass $A$, $x_A$ and that fallen by mass B $x_B$ the equation of motion for each mass is given by:
$$m \ddot x_A=mg+T$$
$$m \ddot x_B=mg-T$$
$T$ is a function of $x_A$ and $x_B$, ($T=k(x_B-x_A-L)$ where $k$ is the spring constant, and $L$ is the natural length) and we cannot assume that $\ddot x_A=g$ or $\ddot x_B=g$. These sorts of equations are called coupled differential equations and can be solved a number of ways.
Best Answer
Of course, the solution is correct. Letting the system fall, the gravitation doesn't act on it anymore, and you have a mass-spring-mass system oscillating in absence of external forces.