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"  $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [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](http://www.math.lsa.umich.edu/~idolga/631.pdf) , Example 13.2