Free body diagram and X1/F1 of two-mass system

Displacements $x_1$ ($M_1$, friction $B_1$, force $f_1$) and $x_2$ ($M_2$, friction $B_2$, spring $K_2$ to the wall). $K_{12} \parallel B_{12}$ couples them.

Free body diagrams

 Opposing forces (to the left)        Applied
 M1x1'', B1x1'
 K12(x1-x2), B12(x1'-x2')   ┌────┐
        ◄───────────────────┤ M1 ├────► f1(t)
                            └────┘
 M2x2'', B2x2', K2x2
 K12(x2-x1), B12(x2'-x1')   ┌────┐
        ◄───────────────────┤ M2 │     (none)
                            └────┘

(For each mass, the applied force acts in the positive direction and all reaction forces oppose it.)

Differential equations

$$\begin{aligned} M_1\ddot x_1 + B_1\dot x_1 + B_{12}(\dot x_1 - \dot x_2) + K_{12}(x_1 - x_2) &= f_1(t) \\ M_2\ddot x_2 + B_2\dot x_2 + K_2x_2 + B_{12}(\dot x_2 - \dot x_1) + K_{12}(x_2 - x_1) &= 0 \end{aligned}$$

Laplace form

$$\begin{aligned} \left[M_1s^2 + (B_1 + B_{12})s + K_{12}\right]X_1 - (B_{12}s + K_{12})X_2 &= F_1 \\ -(B_{12}s + K_{12})X_1 + \left[M_2s^2 + (B_2 + B_{12})s + K_{12} + K_2\right]X_2 &= 0 \end{aligned}$$

Transfer function

By Cramer's rule:

$$\frac{X_1(s)}{F_1(s)} = \frac{M_2s^2 + (B_2 + B_{12})s + K_{12} + K_2}{\Delta}$$ $$\Delta = \left[M_1s^2 + (B_1+B_{12})s + K_{12}\right]\left[M_2s^2 + (B_2+B_{12})s + K_{12} + K_2\right] - (B_{12}s + K_{12})^2$$

Expanded:

$$\begin{aligned} \Delta =\; & M_1M_2s^4 + \left[B_1M_2 + B_2M_1 + B_{12}(M_1 + M_2)\right]s^3 \\ & + \left[B_1B_2 + B_1B_{12} + B_2B_{12} + K_{12}(M_1 + M_2) + K_2M_1\right]s^2 \\ & + \left[B_1(K_{12} + K_2) + B_2K_{12} + B_{12}K_2\right]s + K_{12}K_2 \end{aligned}$$

Check: at $s = 0$, $X_1/F_1 = (K_{12} + K_2)/(K_{12}K_2) = 1/K_{12} + 1/K_2$, the compliance of the two springs in series.