Relation between Magnitude of Greenhouse Effect and the Concentration of GHGs like $CO_2$
Δ=133.26+0.044[2]

It’s just a numerical coincidence that the 0.044 looks like the molar mass of CO2.  That equation gives 0.044 * 380 = 16 W/m2, which is the reduction in surface downwelling IR if you remove all the CO2 from the present day atmosphere (e.g., see Table 1 of Zhong and Haigh, 2013, Weather, https://doi.org/10.1002/wea.2072, pdf).
Looking at water vapour in the same way exposes some of the limits of the model they’ve set up for this tutorial, which was presumably kept simple just to teach some concepts to their students.  To do the same thing for water vapour as was done for CO2 would require removing 208 W/m2 of greenhouse effect (again, see Table 1 of that article), more than the total greenhouse effect of 150 W/m2 in their model.  The discrepancy is because they have too much solar radiation in the surface energy budget, so the diagnosed greenhouse effect doesn’t have to do as much work to balance a 288 K surface temperature.
A more realistic model would exclude the 80 W/m2 of solar radiation absorbed by the atmosphere, leaving an absorbed solar flux at the surface of 160 W/m2.  The balance is then $390 = 160 + \Delta E$, so $\Delta E = 230$ W/m2 and the revised perturbation equation for CO2 is,
$$\Delta E = 214 + 0.044[CO2]$$
The difficulty with H20 is that, unlike CO2, the concentration of water vapour is highly variable in space and time, so the equivalent equation will depend a lot on what you chose as present-day H2O concentration.  A ball-park figure for mean H2O concentration is 4000 ppm, which leads to the equation,
$$\Delta E = 22 + 0.052[H2O]$$
I should emphasise that this model isn’t intended for any serious calculation.  It’s just a tool to get students used to quantifying parts of the Earth system and how they can trade off against each other.