Tangent space change of bases Let $M\subset \Bbb{R}^m$ be a $k$-dimensional differentiable submanifold. Let $(\varphi, U)$ and $(\psi, V)$ be two charts for $p\in M$ with $\varphi(x)=p$ and $\psi(y)=p$. Then we have two bases for the tangent space $T_p M$, namely $\{ (p, \partial_i \varphi(x)) : i=1,...,k \}$ and $\{ (p, \partial_i \psi(y)) : i=1,...,k \}$.
I would like to find a change of bases from the basis given by $\psi$ to the basis given by $\varphi$.
So we want to express a tangent vector $(p, \partial_i \psi(y))$ as a linear combination of basis elements $(p, \partial_j \varphi(y))$. I'm lacking good ideas for how to approach this, so I'm thankful for any help.
 A: $\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}$Here's the situation for dimension two. The generalization to arbitrary dimension is straightforward.
In classical notation, if $(y_{1}, y_{2})$ are smooth functions of $(x_{1}, x_{2})$, the chain rule reads
\begin{align*}
\frac{\dd}{\dd x_{1}}
  &= \frac{\dd y_{1}}{\dd x_{1}}\, \frac{\dd}{\dd y_{1}}
   + \frac{\dd y_{2}}{\dd x_{1}}\, \frac{\dd}{\dd y_{2}}, \\
\frac{\dd}{\dd x_{2}}
  &= \frac{\dd y_{1}}{\dd x_{2}}\, \frac{\dd}{\dd y_{1}}
   + \frac{\dd y_{2}}{\dd x_{2}}\, \frac{\dd}{\dd y_{2}}.
\end{align*}
In modern guise, if $\phi$ is a diffeomorphism and $y = \phi(x)$, the preceding system expresses the push-forwards $\phi_{*}(\dd x_{i})$ as linear combinations of the $\dd y_{j}$, namely
\begin{align*}
\phi_{*}\frac{\dd}{\dd x_{1}}
  &= \frac{\dd \phi_{1}}{\dd x_{1}}\, \frac{\dd}{\dd y_{1}}
   + \frac{\dd \phi_{2}}{\dd x_{1}}\, \frac{\dd}{\dd y_{2}}, \\
\phi_{*}\frac{\dd}{\dd x_{2}}
  &= \frac{\dd \phi_{1}}{\dd x_{2}}\, \frac{\dd}{\dd y_{1}}
   + \frac{\dd \phi_{2}}{\dd x_{2}}\, \frac{\dd}{\dd y_{2}}.
\end{align*}
Note that the change of basis matrix is just the Jacobian of $\phi$. (If you need to get expressions "in the other direction", push the $\dd y_{j}$ forward by the diffeomorphism $\phi^{-1}$.)

Added: Let $n$, $p$, and $m$ be positive integers, $x = (x_{1}, \dots, x_{n})$ coordinates on $\Reals^{n}$, $y = (y_{1}, \dots, y_{p})$ coordinates on $\Reals^{p}$, and $z = (z_{1}, \dots, z_{m})$ coordinates on $\Reals^{m}$. If $g:\Reals^{n} \to \Reals^{p}$ and $f:\Reals^{p} \to \Reals^{m}$ are smooth mappings, $Dg(x)$ is the $p \times n$ matrix of partial derivatives of $g$ at $x$, and $Df(y)$ is the $m \times p$ matrix of partials of $f$ at $y = g(x)$, the chain rule asserts that
$$
D(f \circ g)(x) = Df\bigl(g(x)\bigr) Dg(x).
$$
Classically, if $y = g(x)$ and $z = f(y)$, the preceding matrix product takes the form
$$
\frac{\dd z_{i}}{\dd x_{j}}
  = \sum_{k=1}^{p} \frac{\dd z_{i}}{\dd y_{k}}\, \frac{\dd y_{k}}{\dd x_{j}}.
\tag{1}
$$
(The standard caveat is: On the left, "$z = (f \circ g)(x)$ is a function of $x$", while on the right, "$z = f(y)$ is a function of $y$".)
As you say, "a tangent vector is a directional derivative"; that means a differential operator, not a partial derivative of a particular mapping. In a chart with coordinates $x$, the coordinate vector fields are the differential operators
$$
\frac{\dd}{\dd x_{1}}, \dots, \frac{\dd}{\dd x_{n}}
$$
associated to the standard basis fields in $\Reals^{n}$. (Incidentally, this means the notation "$(p, \dd_{i}\psi(y))$" is anomalous, because you're not taking partial derivatives of $\psi$. Presumably it means "$V$ is a neighborhood of $p$, $\psi:V \to \Reals^{k}$ is a diffeomorphism onto its image, $y = (y_{1}, \dots, y_{k})$ are the associated local coordinates in $V$, i.e., $y$ is the expression of $\psi$ in local coordinates, and the $\dd_{i} = \dd/\dd y_{i}$ are coordinate vector fields.")
Equation (1) holds for arbitrary smooth functions $z_{i}$, so we may (by "erasing the $z$'s") interpret (1) as a relationship between tangent vector fields. The resulting equation (with $n = p = m = 2$) is the system in my original post.
In your situation, $y = \psi \circ \varphi^{-1}(x)$.
I hope that clarifies while still leaving an educational amount of computation and pondering to you. If the business of differential operators is baffling, do you have an instructor you can consult?
