Let $M$ be a smooth manifold. The tangent space of $M$ at a point $p \in M$ is the set
$$T_pM = \{ v_{\gamma, p} | \text{ for all smooth curves } \gamma \text{ through } p \}.$$
Where $v_{\gamma, p}: C^{\infty}(M) \to \mathbb{R}$, $f \mapsto v_{\gamma, p} f = (f \circ \gamma)'(\lambda_0)$, where $\gamma(\lambda_0) =p$. If we push $v_{\gamma, p}$ into a chart $(U, x)$, the result is not ill-defined, and
\begin{align*}
v_{\gamma, p}f &= (f \circ x^{-1} \circ x \circ \gamma)'(\lambda_0) \\
&= \partial_i (f \circ x^{-1})(x(p)) (x \circ \gamma)^{i'} (\lambda_0) \\
&= \dot{\gamma}^i_x \left( \frac{\partial}{\partial x^i} \right)_p f.
\end{align*}
Which leads me to belive $\{ \partial/\partial x^1, \dots, \partial/\partial x^d \}$ is a basis for $T_pM $. Which means
$$\lambda^i \left( \frac{\partial x^j}{\partial x^i} \right)_p = 0 \implies \lambda^i =0 $$ where $x^j: M \to \mathbb{R}$. Because that is just a notation (not an actual partial derivative),
$$\lambda^i \left( \frac{\partial x^j}{\partial x^i} \right)_p = \lambda^i \partial_i (x^j \circ x^{-1})(x(p)) = 0$$
Then $\lambda^j =0$. However, I'm not sure why $(\partial x^j/\partial x^i)_p = \delta^j_i$ (ignoring the fact that it looks like a partial derivative). Thanks!
Best Answer
$\left(\frac{\partial}{\partial x^i}\right)_p$ can be (and I usually see it defined) as the vector tangent to the curve $\gamma:I\to M$ given in coordinates by
$$(x\circ\gamma)(t)=\left(x^1(p),\ldots,x^{i-1}(p),x^i(p)+t,x^{i+1}(p),\ldots,x^d(p)\right)$$
So, given $f\in C^\infty(M)$,
$$\left(\frac{\partial}{\partial x^i}\right)_pf=\left.\frac{d}{dt}\right|_{t=0}\left(f\circ\gamma\right)(t)=\left.\frac{d}{dt}\right|_{t=0}\left(f\circ x^{-1}\circ x\circ\gamma\right)(t)$$
which is the directional derivative at $x(p)$ of the function $f\circ x^{-1}:\mathbb{R}^d\to\mathbb{R}$ along the $i^\text{th}$ unit vector in $\mathbb{R}^d$, or, in other words, the partial derivative $$\frac{\partial \left(f\circ x^{-1}\right)}{\partial x^i}$$ (I'm abusing notation a little bit, and $x^i$ here is not the function on $U$ but rather the coordinate in $\mathbb{R}^d$). By taking $f$ as $x^j$, the result follows.
Also, I'm not sure what you mean by "pushing $v_{\gamma,p}$ into a chart", but your definition is not ill-defined to begin with. You just need $(U,x)$ to be a chart around $p$ for the expression that comes after to make sense.