I have written a proof for the following question, but I'm not sure if I missed some subtilities or that I made some mistakes in my notation.
Prove that $$K_\mathbb{R} = \left\{ \left( (y_0,\ldots, y_n), [x_0:\ldots:x_n] \right) \in \mathbb{R}^{n+1} \times \mathbb{R}\mathbb{P}^n : \exists \lambda \in \mathbb{R} \forall i: y_i = \lambda x_i \right\}\ $$ is the total space of a vector bundle of rank one over the real projective space $\mathbb{R}\mathbb{P}^n $
Define the projection $$\pi : K_{\mathbb{R}} \to \mathbb{R}\mathbb{P}^{n} :\left( (y_0,\ldots, y_n), [x_0:\ldots:x_n] \right) \mapsto [x_0:\ldots:x_n], $$
which is obviously smooth since it's a projection. Surjectivity is trivial, so let's study the fibers directly instead. $\pi^{-1}\left( \{[z_0:\ldots:z_n] \}\right)$ consists of all points in $K_{\mathbb{R}}$ such that the second coordinate is the projective point $[z_0:\ldots:z_n]$. For such a point $\left( (y_0,\ldots, y_n), [z_0:\ldots:z_n] \right) $ to be in $K_{\mathbb{R}}$, it means that there exists a $\lambda \in \mathbb{R}$ such that $y_i = \lambda z_i$, {P}for all $i$. So the first coordinate of $\pi^{-1}\left( [z_0:\ldots:z_n] \right)$ consists of all sclar multiplications of $(z_0, \ldots, z_m)$. So $$\pi^{-1}\left( \{[z_0:\ldots:z_n]\} \right) = \{ \left( \lambda(z_0,\ldots, z_n), [z_0:\ldots:z_n] \right) \mid \lambda \in \mathbb{R} \}$$.
Since $ \{\lambda(z_0,\ldots, z_n) \}\times \{[z_0:\ldots:z_n]\}$ can simply be identified with $\mathbb{R}$ by mapping $ \lambda(z_0,\ldots, z_n) \times \{[z_0:\ldots:z_n]\}$ to $\lambda$, we have that the fibers are one dimensional vector spaces.
$K_\mathbb{R}$ is a Hausdorff second-countable topological space (topology inherited of combinations of subspace, product and quotient topology). This is not very interesting, we want to find the charts of this.
Suppose we have $p \in \mathbb{R}\mathbb{P}^n $ and let $(\varphi_i, U_i)$ be a chart of $\mathbb{R}\mathbb{P}^n $ (and say $V_i = \varphi_i(U_i)$) like the one of example 1.9 of the lecture notes, such that $U_i$ is an open neighbourhood of $p$. Note that $$W_i = \pi^{-1}(U_i) = \{ \left(\lambda(z_0,\ldots, z_n), [z_0:\ldots:z_n]\right) \mid \lambda \in \mathbb{R}, z_i \neq 0 \}. $$ Now define $$\psi_i:K_{\mathbb{R}} \to \mathbb{R}^n \times \mathbb{R}, \left(\lambda(z_0,\ldots, z_n): [z_0:\ldots:z_n]\right) \mapsto \left(\varphi_i([z_0:\ldots:z_n]),\lambda \right).$$
Since $\varphi_i$ is a chart, it follows $\psi_i: W_i \to V_i \times \mathbb{R}$ is a homeomorhpism (and so is the map $ \lambda(z_0,\ldots, z_n) \times \{[z_0:\ldots:z_n]\} \mapsto \lambda$. It also follows tgat $(\psi_\alpha,W_\alpha)$ and $(\psi_\beta,W_\beta)$ are compatible, since $(\varphi_\alpha,U_\alpha)$ and $(\varphi_\beta,U_\beta)$ are compatible.
It directly follows that $\psi_i: W_i \to V_i \times \mathbb{R}$ is a diffeomorphism. But since $U_i$ and $V_i$ are diffeomorphic, there exists a differomorhpism $\Psi_i:U_i \times \mathbb{R} \to W_i$. (Since it's a diffeormorphism, we just swapped domain and codomain for the notation to be easier). Note that $$ \pi \circ \Psi_i\left([z_0:\ldots:z_n], \lambda \right) = \pi \left( \lambda(z_0,\ldots, z_n), [z_0:\ldots:z_n] \right) = [z_0:\ldots:z_n] = \pi_1 \left( [z_0:\ldots:z_n], \lambda \right), $$ and since $$ \Psi_i|_{\{[z_0:\ldots:z_n]\} \times \mathbb{R}^n }\left([z_0:\ldots:z_n], \lambda \right) = \left( \lambda(z_0,\ldots, z_n), [z_0:\ldots:z_n] \right) $$ we have that $\Psi_i: \{[z_0:\ldots:z_n]\} \times \mathbb{R} \to \pi^{-1}([z_0:\ldots:z_n])$ obviously is a vector space isomorphism. Now we're done.