Q:

What are the equations of motions for parafoils?

I want to use the following equations so as to describe the motion of parafoil
$$\left[ {\begin{array}{*{20}{c}}{\dot x}\\{\dot y}\\{\dot z}\end{array}} \right] = {V_a}\left[ {\begin{array}{*{20}{c}}{\cos ({\gamma _a})cos({\chi _a})}\\{\cos ({\gamma _a})sin({\chi _a})}\\{ - sin({\gamma _a})}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{w_x}}\\{{w_y}}\\{{w_z}}\end{array}} \right],
$$
where ${\gamma _a}$ is a flight path angle, and ${\chi _a}$ is heading angle in horizontal plane.
$${\gamma _a} =  - arcsin \frac{{{V_{az}}}}{{{V_a}}}$$
and also
$${\gamma _a} =   arctan \frac{{{-1}}}{{{L/D}}}$$
The equations for lift force and drag force lool like these
$${\displaystyle L\,=\,{\tfrac {1}{2}}\,\rho \,V_{a}^{2}\,C_{L}\,A}$$
$${\displaystyle D\,=\,{\tfrac {1}{2}}\,\rho \,V_{a}^{2}\,C_{D}\,A}$$
according to paper https://lib.dr.iastate.edu/cgi/viewcontent.cgi?referer=https://www.google.ru/&httpsredir=1&article=1629&context=etd
But why the reference area $A$ and speed is similar for both cases? I think its projection should be rather smaller for drag force than for lift one. 
But then on can make assumption that 
$${\gamma _a} =   arctan \frac{{{-1}}}{{{C_{L}/C_{D}}}}$$
I don't understand why this is valid for parafoil
Anyway, I don't want to sink in rigid body physics and it will be comfortable for me to stay on next level of abstraction.
For my case the maximun lift-to-drag ration
$L/D = 4$ and the only information i know is the area of parafoil.
It's enough to characterize the simple model but i can't describe the change of speed caused by the change of lift-to-drag ratio due-to symmetric motion of control handles (breakes, srtopes? I'm not sure what defenition is correct). What is the equation that is responsible for the dependence of lift-to-drad ratio on symmetric motion of handles? Or is there other way to write down variation of velocty due-to handles lengths that allows not to consider forces? It would be appropriate if you know how it was done in X-38 parafoil.
In my opinion the shoud be the law like this
$${\gamma _a} = {\mathop{\rm artan}\nolimits} \left( {\frac{{ - 1}}{{L/D}}} \right) = f(\delta )$$ where $\delta$ is symmetric lenghth of each of two  control handles. But as I said I have only area of parafoil that is 400 ${m^2}$ and mass a few ton.

A:

I don't want to sink in rigid body physics and it will be comfortable for me to stay on next level of abstraction.

Then I'm afraid you will stay stuck at your current level.
Your first equation is only a coordinate transform for the airspeed vector.
The reference area is just an arbitrary area. It makes sense to use the same area for all force components, and the wing area has been universally agreed for this. You will note that c$_D$ is quite a bit smaller than c$_L$ - why should the area be different?
When the same area and speed is used, you can calculate with coefficients just as if they were the real forces. This makes comparison between different speeds and different aircraft (or parafoils) much easier.
Regarding the effect of pulling on the risers (yes, that's the proper word): This changes both the load distribution and the local shape of the parafoil and is not easily expressed in a simple equation.

