Finding the Transfer Function of Spring Mass Damper System
Intro

M and m have only one degree of freedom; both can move vertically only. The magnetic force directly acts upon the magnet m, not on the mass M.

To demystify the picture a bit it could be helpful to think of the magnet as positioned on the other side of the table. The picture has been drawn in LTSPICE, and that has no arrows. So the closest approximation to an arrow is the output pin, and as those can only point horizontally to the right, the whole picture is rotated \$90^o\$ to the right. For the same reason the arrows '-y' and '-F' point to the right whereas I would have liked to have drawn arrows 'y' and 'F' to the left. Further, the right \$b_1\$ should read \$b_2\$.



Now it is clear that this is a series connection of masses with dynamical elements between them, so we start writing down the motional equations from the right to left, starting with the electrical equation for m first, which will contain V, y and F.
After that we will write the motional equation for m and for M.
As M isn't affected by a magnetic force, this last equation will give us y as a function of x, which will be used in the first equation to relate x to V.

Electrical

The magnetic force and the movement of the magnet are coupled via the voltage across the coil. And because $$e=\alpha \dot y \, , \, F=\beta i \, , \, V- e=Ri+L \dot i $$
and assuming L is independent of y, we have
$$V-e = V - \alpha \dot y = R i + L \dot i = \frac{R}{\beta}F+\frac{L}{\beta}\dot F $$

Now we have \$ y\$ in terms of \$F\$ (and \$V\$), and we can write the equations of motion by adding all forces upon the moving objects and forcing them to be zero (by law).

The magnet

$$ F+m \ddot y + b_2 \left(\dot y - \dot x \right) +k_2 \left(y-x \right) =0 $$
We can solve the relationship above between F and y in the s-domain
$$ V-\alpha \dot y = V(s)-\alpha s y = \left(R+Ls\right) i  = \left(R+Ls\right)F/\beta $$
and therefore 
$$ F=\frac{\beta}{R+Ls}\left(V(s)-\alpha s y\right) $$
The sum of the forces on the magnet is zero, so (and for readability the positions are not yet transformed to the s-domain)
$$\frac{\beta V(s)}{R+Ls}-\frac{\alpha \beta }{R+Ls}sy+m\ddot y +k_2\left(y-x\right) + b_2\left(\dot y-\dot x\right)=0 $$
After transforming to the s-domain this equation looks like
$$ \frac{\beta V(s)}{R+Ls}-\frac{\alpha \beta }{R+Ls}sy+ms^2 y +k_2\left(y-x\right) + b_2s\left( y- x\right)= 0 $$
After re-grouping this becomes
$$ ms^2y+\left(b_2-\frac{\alpha \beta }{R+Ls}\right)sy +k_2y -b_2sx - k_2x =-\frac{\beta V(s)}{R+Ls} $$
Isolating \$x\$ and \$y\$ we get
$$ \left(ms^2+b_2s-\frac{\alpha \beta s}{R+Ls}+k_2\right)y-\left(b_2s+k_2\right)x = -\frac{\beta V(s)}{R+Ls} $$

The moving table

For the moving table the governing equation is
$$ M\ddot x +k_1x+b_1\dot x +k_2\left(x-y\right)+b_2\left(\dot x-\dot y\right)=0  $$
After transforming to the s-domain this equations looks like
$$ Ms^2 x +k_1x+b_1s x +k_2\left(x-y\right)+b_2s\left( x- y\right)=0  $$ 
After re-grouping this becomes
$$ -b_2sy -k_2y +Ms^2x +\left(b_1+b_2\right)sx +\left(k_1+k_2\right)x =0 $$
Isolating \$x\$ and \$y\$ we get
$$ - \left( b_2s+k_2 \right) y + \lbrace Ms^2 + \left( b_1+b_2 \right)s +k_1+k_2 \rbrace x = 0 $$
Rewrite this equation to get y in terms of x.
$$ y = \frac{Ms^2 + \left( b_1+b_2 \right)s +k_1+k_2}{b_2s+k_2}x $$

Ensemble

Put \$y=f(x)\$ from above into the relationship between \$x\$, \$y\$ and \$V\$ for the magnet:
$$  \left[ \left(ms^2+b_2s-\frac{\alpha \beta s}{R+Ls}+k_2\right) \frac{Ms^2 + \left( b_1+b_2 \right)s +k_1+k_2}{b_2s+k_2} - \left( b_2s+k_2 \right) \right] x = -\frac{\beta V(s)}{R+Ls} $$

If we multiply both sides of the equation with \$R+Ls\$ we get

$$ \left[ \lbrace (R+Ls) \left( ms^2+bs+k_2 \right) - \alpha\beta s \rbrace \frac{Ms^2+(b_1+b_2)s+(k_1+k_2)}{b_2s+k_2}-(R+Ls)(b_2s+k_2)\right]x = -\beta V(s) $$

Next we multiply both sides with \$b_2s+k_2\$ and get

$$ \left[ \lbrace (R+Ls) \left( ms^2+bs+k_2 \right) - \alpha\beta s \rbrace  \lbrace Ms^2+(b_1+b_2)s+(k_1+k_2)\rbrace - (R+Ls)(b_2s+k_2)^2 \right]x = - (b_2s+k_2) \beta V(s) $$

From visual inspection it follows that we can expect a transfer function \$ x(s)/V(s) \$ with a maximum order of 1 in the nominator and of 5 in the denominator. It is possible that one zero cancels out with the one pole, but that is speculative and would require some more rewriting to find out.