Pullback metric, coordinate vector fields.. I'm doing this computation on $\mathbb{R}^3$ with cylindrical coordinates $(r, \theta, z)$, (which aren't defined on the whole of $\mathbb{R}^3$, but I don't care about that) and I seem to get a contradiction. The problem is as follows.
Make the following change of coordinates:
$$\xi = r,~~\eta = \theta - z,~~\zeta = z.~~~(1)$$
My problem is simple: express the new coordinate vector fields $(\partial_{\xi}, \partial_{\eta}, \partial_{\zeta})$ in terms of the old ones $(\partial_{r}, \partial_{\theta}, \partial_{z})$.
I'm getting confused because I actually have two ways of doing this, both seem right, but they disagree; so let me explain what I've tried.
1. From $(1)$, basically using the Chain Rule:
$$\partial_{\xi} = \frac{\partial{\xi}}{\partial{r}}\partial_r + \frac{\partial{\xi}}{\partial{\theta}}\partial_{\theta} + \frac{\partial{\xi}}{\partial{z}}\partial_z = \partial_r,~~~~(2)\\ \partial_{\eta} = \partial_{\theta} - \partial_z, \\ \partial_{\zeta} = \partial_z. $$
2. From $(1)$, compute the differentials first:
$$d\xi = dr,~~d\eta = d\theta - dz,~~d\zeta = dz.~~~(3)$$
Now from $(3)$ and the general fact that $dx^i(\partial_j) = \delta_{ij}$, where $i, j = 1, 2, 3$ and $\delta_{ij}$ is the Kronecker delta, we must have
$$\partial_{\xi} = \partial_r,~~\partial_{\eta} = \partial_{\theta},~~\partial_{\zeta} = \partial_z + \partial_{\theta}.~~~~(4)$$
But $(2)$ and $(4)$ disagree about who is $\partial_{\zeta}$, for instance. What is going on??
 A: The problem is that equation (2) is wrong. The chain rule holds for differentials but not for vector fields. It is important to understand that 1-forms aka co-vector fields are dual to vector fields and therefore transform differently under coordinate transformations. Because their behaviour under coordinate changes is much nicer, it is almost always easier to use them for explicit calculations.
Co-vector fields transform as the coordinates transform, what is often called "covariant". More precise, their components transform with the Jacobian of the coordinate change. In practice, you can simply use the chain rule to compute the the coordinate differentials, as you did. After that you can compute the vector field using a linear ansatz, e.g. $\partial_{\xi}=a\partial_r+b\partial_\theta+c\partial_z$ and $a=dr(\partial_\xi)=\frac{\partial r}{\partial\xi}=1$ and $b=c=0$. Here you can see what went wrong in your first calculation. The vector fields are actually connected via the inverse transformation. This is why their transformation behaviour is often called "contravariant".
In general, for arbitrary coordinates $x^i$ and $y^i=y^i(x^1,\dots,x^n)$ we have the rules
$$
dy^i=\sum_j \frac{\partial y^i}{\partial x^j}dx^j,
$$ 
and
$$
\partial_{y^i}=\sum_j dx^j(\partial_{y^i})\partial_{x^j}=\sum_j\frac{\partial x^j}{\partial y^i}\partial_{x^j}.
$$
Alternatively, you can also work in components. Let's write an arbitrary vector field as $V=\sum_i V^i \partial_{x^i} = \sum_i \tilde{V}^i \partial_{y^i}$. Then, plugging in the above transformations, we see that the components have to transform in the opposite way to the coordinate vector fields. I.e.
$$
\tilde V^i=\sum_j \frac{\partial y^i}{\partial x^j} V^j.
$$
A similar statement holds for the components of a form $\alpha=\sum_i\alpha_i dx^i$, which now have to transform "contravariantly":
$$
\tilde \alpha_i=\sum_j \frac{\partial x^j}{\partial y^i} \alpha_j.
$$
Side note: These transformation behaviour is the reason for the index notation (up/down) which I also used here. It is a reminder how the object you wrote down transforms.
