To answer the final question, assume that the total space of the taulogical bundle $E\to G(k,n)$ embeds into $G(k,n+1)$ such that $G(k,n+1)\setminus E$ consists of a single point only. Then $G(k,n+1)$ is the Thom space of $E$. The Thom isomorphism for $E$ shows that $H^\ell(G(k,n+1);\mathbb Z/2)=0$ for $0<\ell<k$. But $G(k,n+1)$ is not orientable, so $H^1(G(k,n+1);\mathbb Z/2)\ne 0$. Hence for $k\ge 2$, no such embedding exists.

To answer the final question, assume that the total space of the taulogical bundle $E\to G(k,n)$ embeds into $G(k,n+1)$ such that $G(k,n+1)\setminus E$ consists of a single point only. Then $G(k,n+1)$ is the Thom space of $E$. The Thom isomorphism for $E$ shows that $H^\ell(G(k,n+1);\mathbb Z/2)=0$ for $0<\ell<k$.

On the other hand, the Grassmannian $G(k,n+1)$ approximates the classifying space $BGL(k,\mathbb R)$, and the map $G(k,n+1)\to BGL(k,\mathbb R)$ is $(n+1)$-connected. Hence, there is always a nontrivial homomorphism $\pi_1(G(k,n+1))\to\pi_1(BGL(k,\mathbb R))\cong\mathbb Z/2$, so $H^1(G(k,n+1);\mathbb Z/2)\ne 0$. But this implies that for $k\ge 2$, no embedding as above exists.