The real 2-dim projective space $\mathbb{R}P^2$ can be covered by the following 3 sets of unoriented lines through the origin un $\mathbb{R}^3$:
$ U_x \doteq $ { all lines not lying in the yz plane}
$ U_y \doteq $ { all lines not lying in the xz plane}
$ U_y \doteq $ { all lines not lying in the xy plane}
The lines $L$ belonging to these patches can be parameterized as
$L \in U_x: (u_1 = \frac{y}{x}, u_2 = z/x)$
$L \in U_y: (u_1 = \frac{x}{y}, u_2 = z/y)$
$L \in U_z: (u_1 = \frac{x}{z}, u_2 = y/z)$
How can I fand the transitions between two of these charts, in order to for example investigate differentiability of $\mathbb{R}P^2$ as a 2-dim manifold? Or would I need to investigate the transition function for two coordinate descriptions of a singly line, that differ only by a scaling factor, for that purpose?
In order to answer this question, to start with it helps to write the parameterizations of $U_x$, $U_y$, $U_z$ as functions, where $U_x$, $U_y$, $U_z$ are the ranges of the functions, and the domains are open subsets of $\mathbb{R}^2$ (in this case, all of $\mathbb{R}^2$).
So for example your parameterization of the set $U_z$ is in the form
$$L \in U_z: (u_1 = \frac{x}{z}, \,\, u_2 = \frac{y}{z}) $$
I can see in this formula that you are parameterizing $U_z$ using all of $\mathbb{R}^2$ with $(u_1,u_2)$ coordinates, but it helps to express this parameterization as an actual function $L_z : \mathbb{R}^2 \to U_z$, as follows: $$L_z(u_1,u_2) = \{(x,y,z) \,\bigm|\, \exists s \in \mathbb{R} \, \text{such that} \, x = u_1 s, \,\, y = u_2 s, \,\, z = s\} $$ where I'm using $s$ as a "parameter along the line".
Similarly for $L_y : \mathbb{R}^2 \to U_y$, except I'm going to use $(v_1,v_2)$ as coordinates for $U_y$ in order to avoid confusion, and I'll use $t$ as the parameter along the line: $$L_y(v_1,v_2) = \{(x,y,z) \,\bigm|\, \exists t \in \mathbb{R} \, \text{such that} \, x = v_1 t, \,\, y= t, \,\, z=v_2 t\} $$ Now let's compute a formula for the overlap map $$L_y^{-1} \circ L_z : L_z^{-1}(U_y \cap U_z) \to L_y^{-1}(U_y \cap U_z) $$ The goal is to rewrite the equation $$(v_1,v_2) = L_y^{-1} \circ L_z(u_1,u_2) $$ as two equations of the form $$v_1 = f(u_1,u_2), \,\, v_2 = g(u_1,u_2) $$ and to get explicit formulas for $f$ and $g$.
To do this take any line $L \in U_y \cap U_z$. There is a unique $(u_1,u_2) \in \mathbb{R}^2$ such that $L = L_z(u_1,u_2)$ as defined above. Similarly, there is a unique $(v_1,v_2) \in \mathbb{R}^2$ such that $L=L_y(v_1,v_2)$. Thus we have $L_z(u_1,u_2) = L = L_y(v_1,v_2)$. Using the formulas above we get $$\{(x,y,z) \,\bigm|\, \exists s \in \mathbb{R} \, \text{such that} \, x = u_1 s, \,\, y= u_2 s, \,\, z = s\}$$ $$ = \{(x,y,z) \,\bigm|\, \exists t \in \mathbb{R} \, \text{such that} \, x = v_1 t, \,\, y= t, \,\, z=v_2t\} $$ We must eliminate $x,y,z,s,t$. Setting $x$ coordinates equal, and similarly $y$ coordinates and $z$ coordinates we get $$u_1 s = v_1 t, \,\, u_2 s = t, \,\, s = v_2 t $$ Using $s=v_2 t$ to eliminate $t$ we get $$u_1 v_2 t = v_1 t, \,\, u_2 v_2 t = t $$ Plugging in $t=1$ we get $$u_1 v_2 = v_1, \,\, u_2 v_2 = 1 $$ Solving we get $$v_1 = \frac{u_1}{u_2}, \,\, v_2 = \frac{1}{u_2} $$