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By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*)$$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$

So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea inspired by differential geometry.

The curve "curve" $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [algebraic geometers consider very short curves!] gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}$$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to a colinear vector $(a_0',...,a_n')$, you also have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2

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By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*)$$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$

So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea inspired by differential geometry.

The curve $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}$$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to $(a_0',...,a_n')$, you have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2