warning: This may be a stupid question, because I am poor at differential euqations
$v_p\in T_p\mathbb{R}^n$ is a linear function over $C_p^\infty (\mathbb{R}^n)$, where $C_p^\infty (\mathbb{R}^n)$ denote the functions over $\mathbb{R}^n$ who are smooth at $p$. And $T_p\mathbb{R}^n$ has basis of $\left\{ \frac{\partial }{\partial x^1 }\bigg|_{p},\dots,\frac{\partial }{\partial x^n}\bigg|_{p} \right\}$, $\frac{\partial }{\partial x^j}\bigg|_{p}x^i=\delta_{ij}$, which forms a dual relationship. Does this suggests $\{x^1,\dots,x^n\}$ is a basis of $C^\infty_p(\mathbb{R}^n)$ ? If so, how does linear space $C^\infty_p(\mathbb{R}^n)$ isomorphic to $T^*_p\mathbb{R}^n$ ?
Good question. No. Your definition of $C_p^\infty(\mathbb{R}^n)$ will need some work. As you've described it, it's the functions that are smooth at $p$, but these functions form an infinite-dimension vector space over $\mathbb{R}$ whereas the tangent space is $n$-dimensional, so they can't possibly be dual. Even in dimension one, and even if we think about germs of functions instead of global functions, you still have different functions $x, x^2, \ldots$ which all evaluate the same at the origin but are different in any neighborhood of it.
So it is true that a tangent vector acts on a smooth function, but there is no duality here. In fact, the dual space to tangent vectors is not functions but differential forms.