Free body diagram and X1/F, X2/F of two-mass system

Displacements $x_1$ ($M_1$) and $x_2$ ($M_2$), both positive to the right; $F(t)$ acts on $M_2$.

Free body diagrams

            M1                          M2
 k1x1    ┌──────┐  k2(x2-x1)  k2(x2-x1) ┌──────┐
 ◄───────┤      ├──────►      ◄─────────┤      ├──► F(t)
 M1x1''  │      │                M2x2'' │      │
 ◄───────┤  M1  │                ◄──────┤  M2  │
 B1x1'   │      │                B2x2'  │      │
 ◄───────┤      │                ◄──────┤      │
         └──────┘                       └──────┘

• On $M_1$: inertia $M_1\ddot x_1$, friction $B_1\dot x_1$, spring $k_1x_1$ (all opposing motion); the spring $k_2$ pulls it to the right with $k_2(x_2 - x_1)$, i.e. a force $k_2(x_1 - x_2)$ opposing the motion.
• On $M_2$: applied $F$; opposing $M_2\ddot x_2$, $B_2\dot x_2$, $k_2(x_2 - x_1)$.

Differential equations

$$\begin{aligned} M_1\ddot x_1 + B_1\dot x_1 + k_1x_1 + k_2(x_1 - x_2) &= 0 \\ M_2\ddot x_2 + B_2\dot x_2 + k_2(x_2 - x_1) &= F(t) \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + B_1s + k_1 + k_2)X_1 - k_2X_2 &= 0 \\ -k_2X_1 + (M_2s^2 + B_2s + k_2)X_2 &= F \end{aligned}$$

Determinant:

$$\Delta = (M_1s^2 + B_1s + k_1 + k_2)(M_2s^2 + B_2s + k_2) - k_2^2$$

Expanded:

$$\begin{aligned} \Delta =\; & M_1M_2s^4 + (B_1M_2 + B_2M_1)s^3 + \left[B_1B_2 + M_1k_2 + M_2(k_1 + k_2)\right]s^2 \\ & + \left[B_1k_2 + B_2(k_1 + k_2)\right]s + k_1k_2 \end{aligned}$$

Transfer functions (Cramer's rule)

$$\frac{X_2(s)}{F(s)} = \frac{M_1s^2 + B_1s + k_1 + k_2}{\Delta}$$ $$\frac{X_1(s)}{F(s)} = \frac{k_2}{\Delta}$$

Check: at steady state ($s\to 0$) with a constant force, $X_1/F = 1/k_1$ and $X_2/F = (k_1+k_2)/(k_1k_2) = 1/k_1 + 1/k_2$, which is correct for two springs in series.