So I'm reading about manifolds, and I'm coming to terms with the definitions and all - more specifically, the tangent space $T_{p}M$ - which is a set of mappings, which map a smooth function through $p$ (more precisely, an equivalence class of functions, but as far as I understand, we still write $f$ instead of $[f]$, like in $L^{p}$ spaces), into a number, and which are linear and satisfy Leibniz's identity.
Now, with that definition, I understand that $(\frac{\partial}{\partial x_{i}})|_{p}$, given a local chart $(U, \phi)$ where $p \in U$, defined by $$(\frac{\partial}{\partial x_{i}})|_{p}(f) = \frac{\partial}{\partial x_{i}}(f \circ \phi^{-1})(\phi(p))$$ are objects which belong to $T_{p}M$.
However, I don't see why this is a basis of $T_{p}M$. Why is this so? Given an element $X \in T_{p}M$ , can anyone write out the specific scalars $c_{1},...,c_{n}$ for which $$X = \sum_{k=1}^{n} c_{i} (\frac{\partial}{\partial x_{i}})|_{p}? $$
This follows from the Taylor's theorem. Since chart maps are local diffeomorphisms, we can assume that $M = \mathbb{R^n}$. Without loss of generality, assume that $p = 0$. Take any $f: \mathbb{R^n} \to \mathbb{R}$. Apply Taylor's theorem to get
$$ f(x_1, \ldots, x_n) = f(0) + \frac{\partial f}{\partial x_1}(0) \cdot x_1 + \ldots + \frac{\partial f}{\partial x_n}(0) \cdot x_n + \sum_{i,j} h_{ij} (x_1, \ldots, x_n) \cdot x_i x_j $$
for some $h_{i,j}$ such that $\lim_{\mathbf{x} \to 0} h_{i,j}(\mathbf{x}) = 0$.
Consider any $X \in T_p M$. We have $$ X(f) = X(f(0)) + \frac{\partial f}{\partial x_1}(0) \cdot X(x_1) + \ldots + \frac{\partial f}{\partial x_n}(0) \cdot X(x_n) + \sum_{i,j} X(h_{ij} (x_1, \ldots, x_n) \cdot x_i x_j). $$
Let $c_i = X(x_i)$. These will be your basis coordinates. It's enough to prove that $X(h_{ij} (x_1, \ldots, x_n) \cdot x_i x_j) = 0$. But this follows from Leibniz property of $X$ (try proving it yourself).