Most Economical Description of Two Lines in $\mathbb{P}^{3}$

The simplest way to describe a line $l\subset \mathbb P^3$ in the problem at hand is to give two distinct points on it, say $a=(a_0:a_1:a_2:a_3)$ and $b=(b_0:b_1:b_2:b_3)$.
If you have a second line $m$ joining $c=(c_0:c_1:c_2:c_3)$ to $d=(d_0:d_1:d_2:d_3)$ the condition that the lines $l,m$ meet (i.e. are not skew) is simply $$\det [a^T b^T c^T d^T]=0$$ where you take the determinant of the matrix whose columns are obtained by transposing the four vectors $a,b,c,d$.

Yes. A simple way to do this amounts to choosing a basis for the $\mathbb{C}^4$.

Case 1: You have two skew lines $L_1, L_2$. Have the first line be spanned by $p_1, p_2$ and the second by $p_3, p_4$. These give a basis for the $\mathbb{C}^4$, relative to which the lines are

$$L_1 = [* : * : 0 : 0], \qquad L_2 = [0 : 0 : * : *].$$

Case 2: You have two intersecting lines. Let the point of intersection be $p$. Pick $q_1, q_2$ so that $L_i$ is spanned by $p,q_i$. Finally, pick some last point $r$ not on the plane spanned by $L_1,L_2$.

In the basis $p,q_1,q_2,r$, the two lines are:

$$L_1 = [* : * : 0 : 0], \qquad L_2 = [* : 0 : * : 0].$$