Your system has 1 input, the force $F$, and two outputs, the position of $m_1$ and $m_2$. You should end with two equations, one which describes the behavior of $F$ to $m_1$ and one which describes the behavoir from $F$ to $m_2$, i.e. $\frac{U_1(s)}{F(s)}$ and $\frac{U_2(s)}{F(s)}$. These two equations then describe the behavoir of the coupled system.
So your initial laplace transformed equations, I believe, look something like $\alpha U_1(s) + F(s) = \beta U_2(s)$ and $\gamma U_1(s) = \beta U_2(s)$, where $\alpha, \beta, \gamma$ are polynomials of $s$ and the system parameters. Then you just first rewrite the last equations to $U_1(s) = \beta/\gamma U_2(s)$ such that you can subsitute it in the first by which you obtain $U_1(s)/F(s)$. Then rewrite to $U_2(s) = \gamma/\beta U_1(s)$ and subsitute again to obtain $U_2(s)/F(s)$.
You can simulate the behavoir simply with Simulink or Matlab itself e.g. step(tf) or lsim(tf) with tf the transfer function $U_1(s)/F(s)$ or $U_2(s)/F(s)$.
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Full solution...
We have the following
$$
m_1\ddot{x}_1 = k_2(x_2 - x_1) + d_2(\dot{x}_2 - \dot{x}_1) - k_1x_1 - d_1\dot{x}_1 \\
m_2\ddot{x}_2 = F - k_2(x_2 - x_1) - d_2(\dot{x}_2 - \dot{x}_1)$$
Lets transfer this to the laplace domain and group the terms together such $\alpha X_1(s) = \beta X_2(s)$ and $\gamma X_2(s) = F + \zeta X_1(s)$. With $\alpha, \beta, \gamma, \zeta$ polynomials of $s$ and the coefficients $m_i,k_i,d_i$. We get
$$
\left(m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)\right)X_1(s) = \left(d_2s + k_2\right)X_2(s) \\
\left(m_2s^2 + d_2s + k_2\right)X_2(s) = F(s) + \left(d_2s + k_2\right)X_1(s) \Leftrightarrow F(s) = \left(m_2s^2 + d_2s + k_2\right)X_2(s) - \left(d_2s + k_2\right)X_1(s)
$$
Rewrite the first equation to $\alpha X_1(s) = \beta X_2(s)$ to $X_1(s) = \frac{\beta}{\alpha}X_2(s)$ and $X_2(s) = \frac{\alpha}{\beta}X_1(s)$ we get
$$
X_1(s) = \frac{d_2s + k_2}{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}X_2(s) \\
X_2(s) = \frac{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}{d_2s + k_2}X_1(s)
$$
Now lets plugin each one such that we retrieve expressions for $\frac{X_1(s)}{F(s)}$ and $\frac{X_2(s)}{F(s)}$. Lets do $\frac{X_1(s)}{F(s)}$ first, we get
$$
F(s) = \left(m_2s^2 + d_2s + k_2\right)\frac{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}{d_2s + k_2}X_1(s) - \left(d_2s + k_2\right)X_1(s) \\
F(s) = \frac{\left(m_2s^2 + d_2s + k_2\right)\left(m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)\right) - \left(d_2s + k_2\right)^2}{d_2s + k_2}X_1(s)
$$
Now as you can see the numerator term is a quiet nasty one... You have to first multiply everything and then group all terms such that you get a polynomial of $s$ and the coefficients. For this reason it is also better now to start filling in the values of the coefficients which will make this job then much easier. If you want a analytic solution you can 1) compute it yourself by multiplying everything and collecting the terms 2) let Matlab do this for you. I choose to do it with Matlab.
syms m_1 m_2 d_1 d_2 k_1 k_2 s;
simple((m_2*s^2 + d_2*s + k_2)*(m_1*s^2 + (d_1 + d_2)*s + (k_1 + k_2)) - (d_2*s + k_2)^2)
The simple command will call several commands which could simplify the equation. In this case you will always need the answer for the collect function because this will group the coefficients. In this case it will return
$$
\kappa = (m_1m_2)s^4 + (d_2m_1 + m_2(d_1 + d_2))s^3 + (m_2(k_1 + k_2) + k_2m_1 - d_2^2 + d_2(d_1 + d_2))s^2 + (k_2(d_1 + d_2) - 2d_2k_2 + d_2(k_1 + k_2))s + k_2(k_1 + k_2) - k_2^2
$$
Hence you will get
$$
\frac{X_1(s)}{F(s)} = \frac{d_2s + k_2}{\kappa}
$$
I hope it is clear now and you can solve $\frac{X_2(s)}{F(s)}$ yourself.
-- edit oh well lets do $X_2(s)$ also... --
$$
F(s) = \left(m_2s^2 + d_2s + k_2\right)X_2(s) - \left(d_2s + k_2\right)\frac{d_2s + k_2}{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}X_2(s) \\
F(s) = \left(m_2s^2 + d_2s + k_2\right)X_2(s) - \frac{(d_2s + k_2)^2}{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}X_2(s)
F(s) = \frac{\left(m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)\right)\left(m_2s^2 + d_2s + k_2\right) - \left(d_2s + k_2\right)^2}{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}X_2(s)
\frac{X_2(s)}{F(s)} = \frac{m_1s^2 + (d_1 + d_2)s + (k_1 + k_2)}{\kappa}
$$
Update due to my not looking at the diagram carefully enough and assuming that both springs had one end fixed.
If a force is applied to the left as shown in the diagram the displacement/velocity/acceleration of the mass is only determined by the value of the force the mass and the spring constant of the right hand mass.
Original answer which answers a different question.
Springs connected to two fixed ends and the mass displaced.
The following assumes that the two ends of the springs with the same spring constant not connected to the mass are fixed and the mass is moved to one side.
It is easier, although no essential, to think of the two springs to be unstretched when the mass is in its static equilibrium position and allowing the springs to be compressed as well as being stretched.
In your diagram if the mass is moved to the right the right hand spring becomes compressed and so exerts a force $F$ on the mass to the left.
The left hand spring is stretched to the right and so exerts a force $F$ on the mass to the left.
So the net force on the mass due to the two springs is $2F$ to the left.
A similar analysis can be done if both springs are stretched when the mass is in the equilibrium position.
Moving the mass to the right means the right hand spring pulls less to the right (equivalent to more pull to the left) and the left hand spring pulls more to the left.
Best Answer
What you have worked out so far is the left hand side of the answer. You have written that the sum of the accelerations is zero, but that is not correct because the system is being driven by the roller moving across the sinusoidal surface. Parameterize in terms of time by writing $x=Ut$, then rewrite your $y_0(x)$ equation as $y_0(t)$. Now that you have the vertical position of the roller as a function of time, you can find its acceleration by taking the derivative twice with respect to time. The result should look comforting. This new acceleration term goes with the force that drives the system.