I will try not to skip too many steps. Assuming a Global coordinate frame at the base and the arm is fully extended along the Y-axis of the base frame.
Since SCARA has four joints, we will create four 6D spatial vectors (screws) ${ξ}_{i}$ with respect to a global coordinate frame. Keep in mind that all spatial vectors are described with respect to the global frame. Also ${ξ}_{i}$ will describe the axis around which rotation takes place and along which translation takes place. Basically, we assume a screw motion for every joint along its Z-axis.
The first three elements of vector ${ξ}_{1}$ (for joint 1) will the linear velocity on the screw axis, let's call it ${v}_{i}$. The last three elements are the rotation of the joint with respect to the global coordinate frame, regardless if there are other joints preceding it. Let's call it ${ω}_{i}$ and it will always be a unit vector.$${ξ}_{ι}=\left[{\begin{matrix}{v}_{i}\\{ω}_{i} \end{matrix}}\right]$$
For the first joint the axis of rotation coincides with the global Z-axis. Therefore ${ω}_{1}$ is $${ω}_{1}=\left[{\begin{matrix}0\\0\\1 \end{matrix}}\right]$$
If you are not sure how I came up with it consider the following. if the z axis of the joint was along the y axis of the global frame then the ${ω}_{1}$ would be $${ω}_{y}=\left[{\begin{matrix}0\\1\\0 \end{matrix}}\right]$$. if it was along the x-axis then it would be $${ω}_{x}=\left[{\begin{matrix}1\\0\\0 \end{matrix}}\right]$$. And of course it was along the z axis but pointing downwards then: $${ω}_{-z}=\left[{\begin{matrix}0\\0\\-1 \end{matrix}}\right]$$.
Now let's focus on ${v}_{1}$. We need a point ${q}_{1}$ that will give us the distance of the joint from the origin. The distance is L1 along the z-axis, therefore: $${q}_{1}=\left[{\begin{matrix}0\\0\\L1 \end{matrix}}\right]$$
We rotate ${q}_{1}$ with the unit angular velocity ${ω}_{1}$ (cross product) and we have $${v}_{1}=\left[-{ω}_{1} \times {q}_{1} \right] =- \left[{\begin{matrix}0\\0\\1 \end{matrix}}\right] \times \left[{\begin{matrix}0\\0\\L1 \end{matrix}}\right] = \left[{\begin{matrix}0\\0\\0 \end{matrix}}\right] $$
Therefore $${ξ}_{1}=\left[{\begin{matrix}{v}_{1}\\{ω}_{1} \end{matrix}}\right]=\left[{\begin{matrix}0\\0\\0\\0\\0\\1 \end{matrix}}\right]$$.
Simillarly we have $${ξ}_{2}=\left[{\begin{matrix}{v}_{2}\\{ω}_{2} \end{matrix}}\right]=\left[{\begin{matrix}L2\\0\\0\\0\\0\\1 \end{matrix}}\right]$$
For the third joint the ${ω}_{3}$ is 0 because it is prismatic and ${v}_{3}$ is the unit vector $${v}_{3}=\left[{\begin{matrix}0\\0\\1 \end{matrix}}\right]$$ making ${ξ}_{3}$ $${ξ}_{3}=\left[{\begin{matrix}{v}_{3}\\0 \end{matrix}}\right]=\left[{\begin{matrix}0\\0\\1\\0\\0\\0 \end{matrix}}\right]$$.
For joint 4 $${v}_{4}=-\left[{\begin{matrix} 0\\0\\1\end{matrix}}\right] \times \left[{\begin{matrix}0\\L2+L3\\L1-d1\end{matrix}}\right] = \left[{\begin{matrix}0\\L2+L3\\0\end{matrix}}\right]$$
$${ξ}_{4}=\left[{\begin{matrix}0\\L2+L3\\0\\0\\0\\1\end{matrix}}\right]$$
Now the spatial vectors ${ξ}_{i}$ describe the axes of rotation of all the joints. Keep in mind that some authors have $v$ and $ω$ switch places. For example the paper you are mentioning. This is fine as long as you are consistent throughout.
This example is being worked out in "A mathematical introduction to robotic manipulation" I have linked in my previous answer on page 88.
These are the basics of the screw theory description. Similarly we can create a wrench 6D vector φ with the linear and the angular (torques) forces. As you can tell by now, screw theory is a method to describe the manipulator (like the DH parameters) and not a mobility formula like the Chebychev-Kutzbach-Grubler method. The paper you are referencing stacks the spatial vectors (it calls them twists) ${ξ}_{i}$ and the wrenches ${φ}{i}$ to create a twist matrix and then analyze their constraint using the union and the reciprocal of the twists. Those are vector functions that the screw theory can take advantage of.
I am not sure if my answer helps you but the paper you are mentioning requires a relatively strong understanding of screw theory and vector algebra to make sense.
Please let me know if you need more clarifications.
Cheers.
