I want to prove that on a smooth cubic surface in $\mathbb{P}^3$ there are exactly $27$ lines and I want to do it in a following way. First, our cubic is given by $F(x,y,z,t)=0$ and this polynomial gives a section of $Sym^3(T)^*$ where $T$ is the tautological bundle over $G(2, 4)$ (just by restricting $F$ to the point of $G(2,4)$). The zeros of this section are exactly the lines contained in our surface.

Now I need to show that this section is generic (then use Poincare duality and finish the proof). We need to check that it's transverse to the zero section and that's where the problem is.

I started with taking local coordinates (planes spanned by $(1,0,x_2,x_3)$ and $(0, 1, y_2, y_3)$), one can write the condition for transversality in terms of the tangent spaces (inversibility of some matrix involving $\frac{\partial F}{dx_3}$ and $\frac{\partial F}{dx_4}$) but I failed to show that the tangent vectors actually span the tangent space. Is there an easy way to do that?