First of all you say "why the point $x$ should intersect two skew lines", but that's not what Shafarevich means: he means that the line $l_x$ intersects those lines.
OK, let's check that $\phi$ really is a well-defined rational map.
Fix a point $x \in X_3$ that doesn't lie on either line $m$ or $m'$. Now think of all the lines through $x$ that intersect $m$: the union of these fills up a plane $\Pi \subset \mathbf P^3$. The plane $\Pi$ intersects the line $m'$ in precisely one point (if it didn't, we would have $m' \subset \Pi$, but then $m$ and $m'$ wouldn't be skew). So there is exactly one line through $x$ that intersects both $m$ and $m'$; this is the line $l_x$ we want. It intersects the plane $L$ in one point, and that point is $\phi(x)$.
Next, why is $\phi$ birational? That is, why does it have an inverse rational map? Well, we just reverse the argument above. Fix a point $p \in L$ that doesn't lie on either $m$ or $m'$. Then just as before, there will be a unique line, call it $\lambda_p$, that passes through $p$ and intersects the two lines $m$ and $m'$.
Now the crux: because $X_3$ is a cubic surface, $\lambda_p$ will intersect $X_3$ in exactly three points (counted with multiplicities). One each of these points will be on $m$ and $m'$; define $\psi(p)$ to be the third point. (Notice: this doesn't work as stated if $\lambda_p$ is tangent to $X_3$ at a point on $m$ or $m'$, but that's ok, because that won't happen for most $p$.) Finally, observe that $\phi$ and $\psi$ are inverse rational maps.