Example $9.1$ from John Lee's introduction to smooth manifolds

Let $$(x,y)$$ be the standard coordinates on $$\Bbb R^2$$, and let $$V= \frac{\partial}{\partial x}$$ be the first coordinate vector field. Then the integral curves of $$V$$ are precisely the straight lines parallel to the $$x-$$axis.

I'm trying to verify this result by computing some values, but I think I have understood something wrong.

In John Lee's book introduction to smooth manifolds he defines a vector field as a map $$V :M \to TM$$ and then he defines the value of the vector field at $$p$$ to be $$V_p = V^i(p) \frac{\partial}{\partial x^i} \bigg|_p.$$ My confusion are these $$V^i$$'s. What are the $$V^i$$'s in my case when $$M=\Bbb R^2$$ and $$V= \frac{\partial}{\partial x}$$? Also shouldn't this $$V_p$$ that Lee defines act on some real valued function as we have the partial derivative operator there?

$$V^{i}\in C^{\infty}(M,\Bbb{R})$$ .
For $$V=\frac{\partial}{\partial x}$$ . $$V^{1}(p)=1\,,\forall p\in M$$ and $$V^{2}(p)=0$$ . They are just two constant smooth real valued functions.
For $$M=\mathbb{R}^{2}$$. Take $$V^{1}(x,y)$$ and $$V^{2}(x,y)$$ to be any two smooth real valued functions. Then $$V_{(x,y)}=V^{1}((x,y))\frac{\partial}{\partial x}+V^{2}(x,y)\frac{\partial}{\partial y}$$ is a smooth vector field.