Let $k$ be a field and $ X = \mathbb{P}^n_k $. I have read that every vector field on $ X $, i.e. every global section of $ \mathcal{T}_X = \mathbf{Hom}_{\mathcal{O}_X}(\Omega^1_{X/k},\mathcal{O}_X) $ is induced by a $ k $-derivation $ E : k(x_0,\dots, x_n) \to k(x_0,\dots, x_n) $ with $ E(x_i) $ being homogeneous linear for all $i$.
Indeed given such a derivation on $ D_+( x_j ) \subset \mathbb{P}^n_k$ we can define a derivation on $$ k[x_0/x_i,\dots,x_n/x_i] \to k[x_0/x_i,\dots,x_n/x_i] \\ \frac{x_i}{x_j} \to \frac{x_j E(x_i) - x_i E(x_j)}{x_j^2} $$ and these glue together. However I don't know exactly how to argue that all vector fields are induced this way.
Given a vector field $V$ and $ f/g \in k(x_0,\dots,x_n)^0 $ ($f,g$ are homogeneous of same degree) let $ A = k[x_0,\dots,x_n,1/g]^0 $. Then $D_+(g) = \text{Spec}(A)$. We have $$ V(D_+(g)) \in \text{Der}_k(A,A) $$
We can see this for all $g$. Hence $V$ is induced by a $k$-derivation of $k(x_0,\dots,x_n)^0$. But why is this one induced by a derivation $ E $ of $ k(x_0,\dots,x_n) $ and why is $E(x_i) $ linear homogeneous?
From the Euler Sequence $$0 \rightarrow \mathcal{O}_{\mathbb{P}^n}\rightarrow \mathcal{O}_{\mathbb{P}^n}^{\oplus n+1}(1) \rightarrow \mathcal{T}\mathbb{P}^n\rightarrow 0$$ and ${\rm H}^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})=\{0\}$ it follows that every global section of $\mathcal{T}\mathbb{P}^n$ comes from a $(n+1)$-uple of linear homogeneous polynomials $(A_0, \dots, A_n)$ modulo the radial $(x_0, \dots, x_n)$ i.e. $$(B_0, \dots, B_n) = (A_0 + cx_0 , \dots, A_n + cx_n) $$ define the same section for every $c\in k$. Therefore, you can define your derivation setting $E(x_i) = A_i$ and vice versa.