The key here is that $\langle L,M\rangle$ here is a hyperplane, three-dimensional in $\mathbb{P}^4$. There is, as you have noted, a single point $x$ where $N$ meets this hyperplane.
Now, the union of all lines through $x$ that intersect $L$ is a 2-d plane in $\langle L,M\rangle$. Similarly, the union of lines through $x$ intersecting $M$ is another 2-d plane in $\langle L,M\rangle$. The intersection of these planes will be a line; it can't be a plane because then $L$ and $M$ would intersect, and a dimension argument says it's at least a line. That line passes through $x$, so it's one of the lines that defines each of the planes, and thus it's the line we seek.
Now, seeing how this fits together, an explicit construction:
Define the lines $L$, $M$, $N$ as follows: $L=\{ax_1+bx_2\mid a,b\in \mathbb{R}\}$, $M=\{cy_1+dy_2\mid c,d\in \mathbb{R}\}$, $N=\{ez_1+fz_2\mid e,f\in \mathbb{R}\}$ where $x_1,x_2,y_1,y_2,z_1,z_2$ are pairs of points on each line, with coordinates in $\mathbb{R}^5\setminus \{0\}$. All of these use the description of $\mathbb{P}^4$ as the quotient of $\mathbb{R}^5$ minus the origin by $\mathbb{R}^*$.
Now, the six 5-tuples $x_1,x_2,y_1,y_2,z_1,z_2$ in $\mathbb{R}^5$ must have a nontrivial linear dependence relation; there are constants $A,B,C,D,E,F$ not all zero so that $Ax_1+Bx_2+Cy_1+Dy_2+Ez_1+Fz_2 = 0$. These constants can be found by Gaussian elimination on the matrix with rows $x_1,x_2,y_1,y_2,z_1,z_2$ augmented with a $6\times 6$ identity.
Since the lines don't lie in a hyperplane, our six points must span $\mathbb{R}^5$, and the constants $A,B,C,D,E,F$ are unique up to constant multiples. Also, since $L$ and $M$ don't intersect, there's no nontrivial linear combination of $x_1,x_2,y_1,y_2$ that's zero - which means that at least one of $E$ and $F$ is nonzero. Similarly, at least one of $A$ and $B$ is nonzero, and at least one of $C$ and $D$ is nonzero.
The line we seek is the set of linear combinations $$\{\alpha(Ax_1+Bx_2)+\beta(Cy_1+Dy_2)+\gamma(Ez_1+Fz_2)\mid (\alpha,\beta,\gamma)\in \mathbb{R}^3\}$$
Since the three vectors $Ax_1+Bx_2$, $Cy_1+Dy_2$, $Ez_1+Fz_2$ are linearly dependent, that's a $2$-dimensional subset of $\mathbb{R}^5$, which projects to a line. It intersects $L$ at $Ax_1+Bx_2$ for $\alpha=1,\beta=0,\gamma=0$ and similarly for the other two lines. We could also eliminate the redundancy by restricting to $\alpha+\beta+\gamma=0$.
Yep, you are correct about why your four cases are all distinct. In fact, they are the only distinct cases: any two configurations in the same case are equivalent. To prove this, I find it convenient to pull everything back to $k^5$ (where $k$ is the field) and just think about linear subspaces of $k^5$ using linear algebra.
So, for instance, suppose we are in your case (2a) (similar arguments can be made in your other three cases). Then $\ell_1$ and $\ell_2$ correspond to two-dimensional subspaces $V_1,V_2\subseteq k^5$ with trivial intersection, and $P$ corresponds to a 1-dimensional subspace $W$ which is contained in $V_1+V_2$ but not contained in $V_1$ or $V_2$. Pick a nonzero vector $w\in W$, and write $w=w_1+w_2$ for $w_1\in V_1$ and $w_2\in V_2$ (since $w\in W\subseteq V_1+V_2$). Since $W$ is not contained in $V_1$ or $V_2$, both $w_1$ and $w_2$ are nonzero. So, we can pick $v_1\in V_1$ and $v_2\in V_2$ such that $\{v_1,w_1\}$ and $\{v_2,w_2\}$ are bases for $V_1$ and $V_2$. Finally, pick a nonzero vector $u\in k^5\setminus (V_1+V_2)$, so $\{v_1,w_1,v_2,w_2,u\}$ is a basis for $k^5$.
Now, given any other configuration $(\ell_1',\ell_2',P')$ which is also in case (2a), we can similarly pick a basis $\{v_1',w_1',v_2',w_2',u'\}$. There is then a linear isomorphism $T:k^5\to k^5$ mapping $v_1$ to $v_1'$, $w_1$ to $w_1'$, and so on. This $T$ will then map $V_1$ to $V_1'$, $V_2$ to $V_2'$, and $W$ to $W'$, so it induces a homography that maps $\ell_1$ to $\ell_1'$, $\ell_2$ to $\ell_2'$, and $P$ to $P'$.
Best Answer
It's simple. $L,M,N$ are pairwise non-intersecting if each pair doesn't intersect; that is, $L\cap M$, $M\cap N$, and $N\cap L$ are all empty.
This principle applies to many other "pairwise X" phrases as well.