There's two kinds of $k$-planes in $\mathbb{R}^n\times \mathbb{R}$: Those that project isomorphically to $\mathbb{R}^n$, and those that contain $\mathbb{R}$.

The former are given as graphs of linear functions $V\rightarrow \mathbb{R}$, where $V\subset \mathbb{R}^n$ is a $k$-plane. With the canonical scalar product on $\mathbb{R}^n\times\mathbb{R}$ restricted to $V$, linear functions on $V$ naturally correspond to vectors. This means that we just constructed a bijection between tuples $(V,v), V\subset \mathbb{R}^n, v\in V$ and an open subset of $Gr(k, n+1)$. The former is the usual description of the tautological bundle on $Gr(k, n)$.

The complement (the planes that intersect $\mathbb{R}$) is in natural bijection to $Gr(k-1,n)$, so we can see that that is your complement. In particular, you'll only get a point for $k=1$, the $\mathbb{R}P^n$-case.

There's two kinds of $k$-planes in $\mathbb{R}^n\times \mathbb{R}$: Those that project isomorphically to $\mathbb{R}^n$, and those that contain $\mathbb{R}$.

The former are given as graphs of linear functions $V\rightarrow \mathbb{R}$, where $V\subset \mathbb{R}^n$ is a $k$-plane. With the canonical scalar product on $\mathbb{R}^n\times\mathbb{R}$ restricted to $V$, linear functions on $V$ naturally correspond to vectors. This means that we just constructed a bijection between tuples $(V,v), V\subset \mathbb{R}^n, v\in V$ and an open subset of $\operatorname{Gr}(k, n+1)$. The former is the usual description of the tautological bundle on $\operatorname{Gr}(k, n)$.

The complement (the planes that intersect $\mathbb{R}$) is in natural bijection to $\operatorname{Gr}(k-1,n)$, so we can see that that is your complement. In particular, you'll only get a point for $k=1$, the $\mathbb{R}P^n$-case.