Generic section of a vector bundle over G(2,4) 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?
 A: Your question is beafully answered in Harris' paper Galois Groups of Enumerative Problems, Duke Math. J. Volume 46, Number 4 (1979), 685-724.
More precisely, in Section III Harris proves the following result:

Proposition. Let $V \subset \mathbb{P}^n$ be a smooth hypersurface and $Z(V)$ the Fano scheme of lines on $V$. For any $u \in Z(V)$ let us denote by $l_u$ the line in $V$ corresponding to $u$. Then $Z(V)$ is singular at $u$ if and only if the normal bundle $N$ of $l_u$ in $V$ is special, that is $h^1(V, N) >0$.

He works in in local coordinates, exactly as you attempted to do.   
Let now $V \subset \mathbb{P}^3$ be an irreducible cubic surface. It is not difficult to see that if $V$ contains infinitely many lines then it must be singular. Hence if $V$ is smooth the Fano scheme $Z(V)$ is zero-dimensional.  
By adjunction any line $l$ on a cubic surface $V$ is a $(-1)$-curve, hence $N=\mathcal{O}_{\mathbb{P}^1}(-1)$. Therefore $h^1(V, N)=0$, i.e. any line on $V$ correspond to a smooth point of $Z(V)$. 
This precisely means that the section of $\textrm{Sym}^3 T^*$ whose zero locus is $Z(V)$ is transverse to the zero section. Hence the number of lines on $V$coincides with the degree of the zero-dimensional scheme $Z(V)$, which is given by  $$c_4(\textrm{Sym}^3 T^*)=27.$$
