I think your diagram is missing an angle for the laser angle with respect to the vehicle body -- I'm going to call that angle $\alpha$, see this diagram for clarity:
Since it seems you are tracking an object with your laser, I imagine the point of this is to predict the angular velocity of that object in the vehicle frame. So your laser is scanning in 2D, providing range at a number of angles, and you want to predict what angle the object will be at in the next time step.
In that case, you need to consider to motion of the system to figure out the motion of the target object in the laser frame. Assume the target object has position $\hat{p}$ in the world frame and $\hat{p}_L$ in the laser frame. Although it sounds like this is for a manipulator that is fixed at the base, we will still consider a moving centre of mass with position $p_0$ in the world frame. The position of the target in the world frame is then:
$\hat{p} = p_0 + A_{\beta} \begin{bmatrix} L_z \\ -L_y \\ \end{bmatrix} + A_{\beta} R \begin{bmatrix} \cos \alpha \\ -\sin \alpha \\ \end{bmatrix} $
Where $A_{\beta} = \begin{bmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \\ \end{bmatrix}$
Then we differentiate that to get the velocity -- which should be zero assuming your object is stationary in the world frame.
$0 = \dot{p}_0 + \dot{\beta} J_{\beta} \begin{bmatrix} L_z \\ -L_y \\ \end{bmatrix} + R \dot{\beta} J_{\beta} \begin{bmatrix} \cos \alpha \\ -\sin \alpha \end{bmatrix} + R \dot{\alpha} A_{\beta} \begin{bmatrix} -\sin \alpha \\ -\cos \alpha \end{bmatrix} + \dot{R} A_{\beta} \begin{bmatrix} \cos \alpha \\ -\sin \alpha \end{bmatrix}$
Where $J_{\beta} = \begin{bmatrix} -\sin \beta & \cos \beta \\ -\cos \beta & -\sin \beta \\ \end{bmatrix}$
This then defines two equations (the z- and y-components of the above equation) in two unknowns ($\dot{R}$ and $\dot{\alpha}$).
However, what you really need is to simply find an expression for the predicted range and laser angle ($\tilde{R}$ and $\tilde{\alpha}$) in terms of your states, the vehicle pose, ($z$,$y$,$\beta$) and target position ($\hat{z}$,$\hat{y}$).
$\tilde{R} = \sqrt{\left( \hat{z} - \left( z + L_z \cos \beta \right) \right)^2 + \left( \hat{y} - \left( y + L_y \sin \beta \right) \right)^2}$
$\tilde{\alpha} = \tan^{-1} \left( \frac{\hat{y} - \left( y + L_y \sin \beta \right)}{\hat{z} - \left( z + L_z \cos \beta \right) } \right) - \beta$
Then you can apply this in an EKF according to the chat discussion. Note I have switched the axes labels and sign conventions to something that I am more used to.
