Let $i: \mathbb{P}^1 \to \mathbb{P}^2$ denote the inclusion $[x_0,x_1] \mapsto [x_0,x_1,0]$. In other words, we identify $\mathbb{P}^1$ as the subvariety defined by ${x_2 = 0}$ in $\mathbb{P}^2$. The inclusion induces the following short exact sequence $$0 \to T_{\mathbb{P}^1} \to i^*T_{\mathbb{P}^2} \to N_{\mathbb{P}^1/\mathbb{P}^2} \to 0,$$ where the normal bundle $N_{\mathbb{P}^1/\mathbb{P}^2} \cong \mathcal{O}_{\mathbb{P}^2}(1)|_{\mathbb{P}^1}$. Then we have a long exact sequence of sheaf cohomology $$0 \to H^0(\mathbb{P}^1,T_{\mathbb{P}^1}) \to H^0(\mathbb{P}^1,i^*T_{\mathbb{P}^2})\to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^2}(1)|_{\mathbb{P}^1}) \to H^1(\mathbb{P}^1,T_{\mathbb{P}^1}) \to \cdots.$$
It's known that $H^1(\mathbb{P}^n,T_{\mathbb{P}^n}) = 0$ in general. So the above sequence reduces to $$0 \to H^0(\mathbb{P}^1,T_{\mathbb{P}^1}) \to H^0(\mathbb{P}^1,i^*T_{\mathbb{P}^2})\to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^2}(1)|_{\mathbb{P}^1}) \to 0.$$
I want to describe the above short exact sequence explicitly over the affine open set $U_0 = \{[x_0 \neq 0,x_1]\}$ with coordinate $x = x_1/x_0$.
I know $H^0(\mathbb{P}^1,T_{\mathbb{P}^1})$ has a basis of three vector fields $$\frac{\partial}{\partial x}, x \frac{\partial}{\partial x}, x^2 \frac{\partial}{\partial x}.$$ I also know $H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^2}(1)|_{\mathbb{P}^1})$ has a basis $x_0,x_1$ ($x_2$ is not one of them because it vanishes on $\mathbb{P}^1$). Over $U_0$, they become two functions $$1,x.$$ But I'm not sure how to write global sections from $H^0(\mathbb{P}^1,i^*T_{\mathbb{P}^2})$.
How to identify $i^*T_{\mathbb{P}^2}$ as a vector bundle of rank $2$ on $\mathbb{P}^1$ so we can write down a basis of $H^0(\mathbb{P}^1,i^*T_{\mathbb{P}^2})$?
I worked out the details. They were added in an edit of my question. For the benefit of others, I post them here as a complete answer.
To describe an element in $H^0(\mathbb{P}^1, i^*T_{\mathbb{P}^2})$, we consider $U_0 \cong \mathbb{C}$ of $\mathbb{P}^1$ with coordinate $x = x_1/x_0$ sitting inside $V_0 \cong \mathbb{C}^2$ of $\mathbb{P}^2$ with coordinates $$x = x_1/x_0, \ \ \ y = x_2/x_0.$$
With this change of variables, the following $9$ vector fields on $\mathbb{C}^3$
$$x_0 \frac{\partial}{\partial x_0}, x_1 \frac{\partial}{\partial x_0},x_2 \frac{\partial}{\partial x_0}, x_0 \frac{\partial}{\partial x_1}, x_1 \frac{\partial}{\partial x_1}, x_2 \frac{\partial}{\partial x_1}, x_0 \frac{\partial}{\partial x_2}, x_1 \frac{\partial}{\partial x_2}, x_2 \frac{\partial}{\partial x_2}$$
are descended to the following vector fields on $V_0$
$$-x \partial_x - y \partial_y,\ -x^2 \partial_x - xy \partial_y,\ -xy \partial_x - y^2 \partial_y,\ \partial_x,\ x \partial_x,\ y \partial_x,\ \partial_y,\ x \partial_y,\ y \partial_y$$
respectively. The first one is redundant since $$(-x \partial_x - y \partial_y) + x \partial_x + y \partial_y = 0.$$
Since $y = 0$ on $U_0$, we also discard $$-xy \partial_x - y^2 \partial_y,\ y \partial_x,\ y \partial_y.$$ So we obtain the following basis of $H^0(\mathbb{P}^1, i^*T_{\mathbb{P}^2})$:
$$-x^2 \partial_x - xy \partial_y,\ \partial_x,\ x \partial_x,\ \partial_y,\ x \partial_y$$
Now the map $H^0(\mathbb{P}^1, T_{\mathbb{P}^1}) \to H^0(\mathbb{P}^1, i^*T_{\mathbb{P}^2})$ sends $$\partial_x \mapsto \partial_x,\ x\partial_x \mapsto x\partial_x,\ x^2\partial_x \mapsto x^2\partial_x + xy\partial_y,$$
and the next map $H^0(\mathbb{P}^1, i^*T_{\mathbb{P}^2}) \to H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^2}(1)|_{\mathbb{P}^1}) $ sends $$f(x,y)\partial_x + g(x,y)\partial_y \mapsto g(x,0).$$
One can check these two maps do give the exactness at $H^0(\mathbb{P}^1, i^*T_{\mathbb{P}^2})$.
Remark. The above works without identifying the vector bundle $i^*T_{\mathbb{P}^2}$. As mentioned by Sasha, the original normal bundle sequence splits and hence $i^*T_{\mathbb{P}^2} \cong \mathcal{O}(1)\oplus\mathcal{O}(2)$ over $\mathbb{P}^1$. One can see that the global sections of $\mathcal{O}(1)$ correspond to $\partial_y,\ x \partial_y$ and the global sections of $\mathcal{O}(2)$ correspond to $-x^2 \partial_x - xy \partial_y,\ \partial_x,\ x \partial_x$.