Free body diagram of masses between walls

Displacements $x_1$ ($M_1$, force $f$) and $x_2$ ($M_2$); no ground friction. $K_1$ ties $M_1$ to the left wall, $B$ couples the masses and $K_2$ ties $M_2$ to the right wall.

Free body diagrams

          M1                            M2
 K1x1  ┌──────┐ f(t)    B(x2'-x1') ┌──────┐
 ◄─────┤      ├──►      ◄──────────┤      │
 M1x1''│  M1  │         M2x2''     │  M2  │
 ◄─────┤      │         ◄──────────┤      │
 B(x1'-x2')   │         K2x2       │      │
 ◄─────┤      │         ◄──────────┤      │
       └──────┘                    └──────┘

Differential equations

$$\begin{aligned} M_1\ddot x_1 + K_1x_1 + B(\dot x_1 - \dot x_2) &= f(t) \\ M_2\ddot x_2 + K_2x_2 + B(\dot x_2 - \dot x_1) &= 0 \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + Bs + K_1)X_1 - BsX_2 &= F \\ -BsX_1 + (M_2s^2 + Bs + K_2)X_2 &= 0 \end{aligned}$$

From the second equation $X_2 = \dfrac{Bs}{M_2s^2 + Bs + K_2}X_1$. Substituting:

$$\left[(M_1s^2 + Bs + K_1) - \frac{B^2s^2}{M_2s^2 + Bs + K_2}\right]X_1 = F$$

Transfer function

$$\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + Bs + K_2}{(M_1s^2 + Bs + K_1)(M_2s^2 + Bs + K_2) - B^2s^2}$$

Expanding the denominator:

$$\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + Bs + K_2}{M_1M_2s^4 + B(M_1 + M_2)s^3 + (M_1K_2 + M_2K_1)s^2 + B(K_1 + K_2)s + K_1K_2}$$

Check: for a steady force ($s\to 0$), $X_1 = F/K_1$; the damper carries no static force, so $K_2$ does not share the load.