Say the four points are $(x_1, y_1)\ldots (x_4, y_4)$.
You want $a,b,c,d$ to satisfy the four equations: $$y_1 = ax_1^3 + bx_1^2 + cx_1 + d\\
y_2 = ax_2^3 + bx_2^2 + cx_2 + d\\
y_3 = ax_3^3 + bx_3^2 + cx_3 + d\\
y_4 = ax_4^3 + bx_4^2 + cx_4 + d$$
Do you know how to solve four linear equations in four unknowns?
I will work an example, since it is not hard to pick up. Let's say we have points $(1,3), (2,9), (3,27), (4,63)$. Then we get the four equations:
$$\begin{eqnarray}
3 &=& a + b + c + d\\
9 &=& 8a + 4b + 2c + d\\
27 &=& 27a+ 9b + 3c + d\\
63 &=& 64a + 16b + 4c + d
\end{eqnarray}$$
We start by subtracting each equation from the following equation, which eliminates all the $d$'s:
$$\begin{eqnarray}
6 &=& 7a + 3b + c \\
18 &=& 19a + 5b + c \\
36 &=& 37a+ 7b + c
\end{eqnarray}$$
Again we subtract each equation from the following one, which eliminates the $c$'s:
$$\begin{eqnarray}
12 &=& 12a + 2b \\
18 &=& 18a + 2b
\end{eqnarray}$$
We could subtract again, giving $6a=6$, so $a=1$, or
we can solve this pair of equations by inspection: evidently $a=1$ and $b=0$.
Then we substitute $a=1$ and $b=0$ back into one of the previous batch of equations, say $6 = 7a + 3b + c$, giving $6 = 7 + c$ and then $c=-1$.
Then we substitute $a=1, b=0, c=-1$ into one of the original equations, say $3 = a + b + c + d$, giving $3 = 1-1+d$ and so $d=3$.
Now we have $a=1, b=0, c=-1, d=3$, so the cubic polynomial we found is $y= x^3 -x + 3$, which is correct.
Your lines must intersect to determine a plane; skew lines don't define a plane. But if they do intersect, then produce 3 non-collinear points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$. In general, you will have a planar equation of the form $z = ax + by + c$ (except for the special case of a vertical plane). Your coefficients are the 3 unknowns, and you have three points. Using these 3 points, you can solve for your coefficients and get an equation for your plane.
Since you have two different lines, you can choose any two points on one of the lines, and a third point on the other line; you do not need to use their point of intersection. For convenience, try taking as many of the coordinates to be zero as possible.
In practical applications, your data will not be this nice. You will have a number of points scattered throughout space, not all lying on the same plane. Every three of these determines a plane, so it's typical to form a triangular mesh by repeatedly applying the above method for each local triplet. It's also more common to form two vectors and take a cross product to get a normal to the plane, and to write it in point-normal form; you may want to look into that method after you're comfortable with the one I mentioned above, especially if you're already familiar with vectors and the cross product.
Update: Now that I see the actual situation you have, your question is not basic at all. In fact, you have almost no hope of getting what I think you want out of your data. Your interpolated guess in the second figure is not likely close to the truth, but it also not the worst thing you could do with your data. I don't know what algorithm you used (or your software used) to get the interpolation, but there are other ways to guess the value of your function at the missing points. You should read the various methods outlined at http://en.wikipedia.org/wiki/Multivariate_interpolation. This is not my area of expertise, so maybe someone else will weigh in, but I think that you can probably find a method there that produces results that you prefer to the interpolation you already have, and also with some parameters that let you fine-tune the results. I think you'll probably end up with, at best, a very bad guess for the missing data, especially if you want to interpolate far from either of the curves. You will also need to choose a discrete subset of your curves to employ most of the methods, but I think you probably already have a discrete dataset that you used to produce them in the first place.
Best Answer
In order to completely define a cubic function, you need four distinct points. So it is not very surprising to me that you are getting these conflicting results, since two points leaves a great deal of ambiguity.
That said, if you know in advance your equation looks like $y=ax^3+b$, then two points is enough to specify it. [Note: when I say "looks like" this I mean it has no $x$ or $x^2$ terms].
The general method for finding the coefficients in problems like this is to plug in the points for $x$ and $y$. This leaves a linear system of equations in the coefficients, which is then solvable by standard methods (like substitution or elimination).
In your case, we have the $y$-intercept $(0,2)$ and the root $(4,0)$, which gives $$0=a(4)^3+b$$ $$2=a(0)^3+b$$
The second equation immediately gives $b=2$, and then by substituting into the first we see that $0=64a+2$ and so $a=-\frac{1}{32}$.
I'm not sure where you went wrong precisely because you didn't show your reasoning all the way through [you did write the $y$-intercept as $(2,0)$ instead of $(0,2)$, though I don't think that was the real problem]. But hopefully with this solution you can diagnose the problem yourself :)