As mentioned in the comments, this is an extremely complex problem if you intend to consider every possible aspect. However, for a general estimation, you can use the relatively simple methods described in this document to begin calculating the effects of air drag on projectiles.
Note that in the document cited, they make the assumption that the air is not moving, and begun their derivation from $f = Dv^2$, and this $v$ was relative to the air and therefore the following equations simply used the velocity of the ball. For the more complex case where the air is moving as well, you will need to account for this change and make sure that the x and y components of the force due to drag are calculated using the relative velocity of the projectile through the now-moving air.
Also worth noting is the fact that if the wind direction changes, the effective footprint of your projectile will change, thus changing $D$ and therefore the force due to drag. If you are willing to make a reasonable approximation for the average footprint of your projectile, however, this will likely yield a result that is accurate enough for your purposes.
Hope this helps!
You said this is a one dimensional problem, and the drag force is proportional to the square of the velocity $v$, so
$$
F = - b v^2 = m \frac{dv}{dt} \ \ \ \Rightarrow \ \ \ \int_{v\left(0\right)}^{v\left(t\right)}\frac{dv}{v^2} = - \frac{b}{m} \int_0^t dt'
$$
The above integral (I'll let you do it, or see this) gives $v\left(t\right)$. The average velocity (or speed, since $v\left(t\right)>0$) from $t=t_1$ to $t=t_2$ is then
$$
\left<v\left(t\right)\right>_{12} = \frac{1}{t_2-t_1} \int_{t_1}^{t_2} dt \ v\left(t\right)
$$
Best Answer
First be aware that, so far, you have been dealing with projectile velocity as a 2-dimensional phenomenon. This is going to have to change.
In order to conform to standard notation, you'll need to refer to altitude (vertical motion) as the z-coordinate, while horizontal will be handled by x and y coordinates. For ease of calculation, you can assume that you initial velocity has a zero component in one axis, let's say the y axis. Now the equations you're familiar with look the same, except that they are in terms of x and z rather than x and y. Let's start by looking at how to do the simulation in the absence of crosswind.
At any point, you can calculate the x and z velocities $V_x$ and $V_z$, and combine them to get the speed of the projectile, V: $$V = \sqrt{(V_x)^2+(V_y)^2}$$
Use this to calculate the drag on the projectile from the answer you linked to, then decompose the drag force into its x and z components. Apply the components to modify the x and z velocities and calculate the next step.
Now for crosswind. The simplest approach is to assume that the crosswind is much less than the projectile velocities. In this case, you can assume that velocities in the y axis don't affect the other 2 axes.
Start by calculating the relative crosswind - that is the speed of the wind minus the actual y-axis speed. In order to do this, you need to calculate the component of the crosswind which is actually perpendicular to the projectile's motion. For instance, if the wind is at 45 degrees to the projectile's direction, the crosswind sideways to the projectile will be .707 times the wind speed. Recognize that the coefficient of drag is different for crosswind than for the previous calculation, since the "regular" calculation looked at the projectile head-on, while crosswind applies sideways. Knowing the relative crosswind and the crosswind drag coefficient, use the linked answer to calculate the sideways force on the projectile, then apply this to calculate the new velocity and then the new y-axis position.
From the correction for wind direction, it should be clear that, for best accuracy, you ought to factor in the "head-on" component of the wind into the overall drag calculation, but you can always point out that the crosswind is assumed to be much lower than the projectile speed.