Here is how it works for the (complex) Grassmannian. I will leave you the pleasure to extend this point of view to others homogeneous spaces (for instance complete and incomplete flag manifolds).

Firs of all, let me give a slightly different point of view for the Euler exact sequence on the projective space. Let $V$ be a complex vector space of dimension $n+1$ and $P(V)$ its corresponding projective space of lines. Call $\underline V=P(V)\times V$ the trivial rank $n+1$ vector bundle with fiber $V$ on $P(V)$. Then you have the short exact sequence \begin{equation} 0\to\mathcal O(-1)\to\underline V\to H\to 0, \end{equation} where $\mathcal O(-1)$ is the tautological line bundle on $P(V)$ and $H$ is the quotient rank $n$ vector bundle $\underline V/\mathcal O(-1)$.

Now, looking at the differential $d\pi_x\colon V\to T_{P(V),[x]}$ of the projection $\pi\colon V\setminus\{0\}\to P(V)$, it is straightforward to see that it defines a canonical isomorphism $$ T_{P(V)}\simeq H\otimes\mathcal O(1). $$ Tensoring the short exact sequence above by $\mathcal O(1)$ and using this canonical isomorphism you get the Euler exact sequence $$ 0\to \mathcal O\to\underline V\otimes\mathcal O(1)\to T_{P(V)}\to 0 $$ (to get your version just notice that $\underline V\otimes\mathcal O(1)\simeq\mathcal O(1)^{\oplus(n+1)}$, then take the dual exact sequence).

Now, let $V$ be of dimension $d$ and $G_r(V)$ be the Grassmannian of $r$-codimensional vector subspaces of $V$. Consider the tautological subbundle $S\subset \underline V=G_r(V)\times V$ and the associated short exact sequence $$ 0\to S\to\underline V\to Q\to 0, $$ where $Q=\underline V/S$. Similarly as before, it is straightforwardly seen that there is a canonical isomorphism $$ T_{G_r(V)}\simeq S^*\otimes Q\simeq\operatorname{Hom}(S,Q). $$ Tensoring by $S^*$, you get your Euler sequence for the Grassmannian, that is $$ 0\to S^*\otimes S\to\underline V\otimes S^*\to T_{G_r(V)}\to 0. $$ Of course, it reduces to the standard one whenever $r=d-1$, and playing with the short exact sequence you can give it your favorite shape (taking duals, tensoring with other vector bundles...).

Here is how it works for the (complex) Grassmannian. I will leave you the pleasure to extend this point of view to others homogeneous spaces (for instance complete and incomplete flag manifolds).

Firs of all, let me give a slightly different point of view for the Euler exact sequence on the projective space. Let $V$ be a complex vector space of dimension $n+1$ and $P(V)$ its corresponding projective space of lines. Call $\underline V=P(V)\times V$ the trivial rank $n+1$ vector bundle with fiber $V$ on $P(V)$. Then you have the short exact sequence \begin{equation} 0\to\mathcal O(-1)\to\underline V\to H\to 0, \end{equation} where $\mathcal O(-1)$ is the tautological line bundle on $P(V)$ and $H$ is the quotient rank $n$ vector bundle $\underline V/\mathcal O(-1)$.

Now, looking at the differential $d\pi_x\colon V\to T_{P(V),[x]}$ of the projection $\pi\colon V\setminus\{0\}\to P(V)$, it is straightforward to see that it defines a canonical isomorphism $$ T_{P(V)}\simeq H\otimes\mathcal O(1). $$ Tensoring the short exact sequence above by $\mathcal O(1)$ and using this canonical isomorphism you get the Euler exact sequence $$ 0\to \mathcal O\to\underline V\otimes\mathcal O(1)\to T_{P(V)}\to 0 $$ (to get your version just notice that $\underline V\otimes\mathcal O(1)\simeq\mathcal O(1)^{\oplus(n+1)}$, then take the dual exact sequence).

Now, let $V$ be of dimension $d$ and $G_r(V)$ be the Grassmannian of $r$-codimensional vector subspaces of $V$. Consider the tautological subbundle $S\subset \underline V=G_r(V)\times V$ and the associated short exact sequence $$ 0\to S\to\underline V\to Q\to 0, $$ where $Q=\underline V/S$. Similarly as before, it is straightforwardly seen that there is a canonical isomorphism $$ T_{G_r(V)}\simeq S^*\otimes Q\simeq\operatorname{Hom}(S,Q). $$ Tensoring by $S^*$, you get your Euler sequence for the Grassmannian, that is $$ 0\to S^*\otimes S\to\underline V\otimes S^*\to T_{G_r(V)}\to 0. $$ Of course, it reduces to the standard one whenever $r=d-1$, and playing with the short exact sequence you can give it your favorite shape (taking duals, tensoring with other vector bundles...).

Observe finally that $$ \det S^*\simeq\det Q $$ is the (very) ample line bundle on $G_r(V)$ which gives the Plücker embedding into projective space and whose first Chern class generates $H^2(G_r(V),\mathbb Z)$.

Here is how it works for the (complex) Grassmannian. For the moment I will leave you the pleasure to extend this point of view to others homogeneous spaces (for instance complete and incomplete flag manifolds).

Firs of all, let me give a slightly different point of view for the Euler exact sequence on the projective space. Let $V$ be a complex vector space of dimension $n+1$ and $P(V)$ its corresponding projective space of lines. Call $\underline V=P(V)\times V$ the trivial rank $n+1$ vector bundle with fiber $V$ on $P(V)$. Then you have the short exact sequence \begin{equation} 0\to\mathcal O(-1)\to\underline V\to H\to 0, \end{equation} where $\mathcal O(-1)$ is the tautological line bundle on $P(V)$ and $H$ is the quotient rank $n$ vector bundle $\underline V/\mathcal O(-1)$.

Now, looking at the differential $d\pi_x\colon V\to T_{P(V),[x]}$ of the projection $\pi\colon V\setminus\{0\}\to P(V)$, it is straightforward to see that it defines a canonical isomorphism $$ T_{P(V)}\simeq H\otimes\mathcal O(1). $$ Tensoring the short exact sequence above by $\mathcal O(1)$ and using this canonical isomorphism you get the Euler exact sequence $$ 0\to \mathcal O\to\underline V\otimes\mathcal O(1)\to T_{P(V)}\to 0 $$ (to get your version just notice that $\underline V\otimes\mathcal O(1)\simeq\mathcal O(1)^{\oplus(n+1)}$, then take the dual exact sequence).

Now, let $V$ be of dimension $d$ and $G_r(V)$ be the Grassmannian of $r$-codimensional vector subspaces of $V$. Consider the tautological subbundle $S\subset \underline V=G_r(V)\times V$ and the associated short exact sequence $$ 0\to S\to\underline V\to Q\to 0, $$ where $Q=\underline V/S$. Similarly as before, it is straightforwardly seen that there is a canonical isomorphism $$ T_{G_r(V)}\simeq S^*\otimes Q\simeq\operatorname{Hom}(S,Q). $$ Tensoring by $S^*$, you get your Euler sequence for the Grassmannian, that is $$ 0\to S^*\otimes S\to\underline V\otimes S^*\to T_{G_r(V)}\to 0. $$ Of course, it reduces to the standard one whenever $r=d-1$, and playing with the short exact sequence you can give it your favorite shape (taking duals, tensoring with other vector bundles...).

Here is how it works for the (complex) Grassmannian. I will leave you the pleasure to extend this point of view to others homogeneous spaces (for instance complete and incomplete flag manifolds).

Firs of all, let me give a slightly different point of view for the Euler exact sequence on the projective space. Let $V$ be a complex vector space of dimension $n+1$ and $P(V)$ its corresponding projective space of lines. Call $\underline V=P(V)\times V$ the trivial rank $n+1$ vector bundle with fiber $V$ on $P(V)$. Then you have the short exact sequence \begin{equation} 0\to\mathcal O(-1)\to\underline V\to H\to 0, \end{equation} where $\mathcal O(-1)$ is the tautological line bundle on $P(V)$ and $H$ is the quotient rank $n$ vector bundle $\underline V/\mathcal O(-1)$.

Now, looking at the differential $d\pi_x\colon V\to T_{P(V),[x]}$ of the projection $\pi\colon V\setminus\{0\}\to P(V)$, it is straightforward to see that it defines a canonical isomorphism $$ T_{P(V)}\simeq H\otimes\mathcal O(1). $$ Tensoring the short exact sequence above by $\mathcal O(1)$ and using this canonical isomorphism you get the Euler exact sequence $$ 0\to \mathcal O\to\underline V\otimes\mathcal O(1)\to T_{P(V)}\to 0 $$ (to get your version just notice that $\underline V\otimes\mathcal O(1)\simeq\mathcal O(1)^{\oplus(n+1)}$, then take the dual exact sequence).

Now, let $V$ be of dimension $d$ and $G_r(V)$ be the Grassmannian of $r$-codimensional vector subspaces of $V$. Consider the tautological subbundle $S\subset \underline V=G_r(V)\times V$ and the associated short exact sequence $$ 0\to S\to\underline V\to Q\to 0, $$ where $Q=\underline V/S$. Similarly as before, it is straightforwardly seen that there is a canonical isomorphism $$ T_{G_r(V)}\simeq S^*\otimes Q\simeq\operatorname{Hom}(S,Q). $$ Tensoring by $S^*$, you get your Euler sequence for the Grassmannian, that is $$ 0\to S^*\otimes S\to\underline V\otimes S^*\to T_{G_r(V)}\to 0. $$ Of course, it reduces to the standard one whenever $r=d-1$, and playing with the short exact sequence you can give it your favorite shape (taking duals, tensoring with other vector bundles...).