It is not a contradiction at all, and goes back to the well-known and often-emphasized fact that partial derivatives depend on which coordinates you keep constant.
In general, if you have a function $f:M\to\Bbb{R}$ (i.e not your $f(r)$ particularly, but literally any smooth function on any smooth manifold), then you can define its exterior derivative $df$. So, the data of a single function $f$ immediately allows you to define a 1-form $df$. However, a single function $f$ does NOT (barring trivial cases like 1D manifolds with $df$ nowhere vanishing) in general give rise to a vector field $\frac{\partial}{\partial f}$. Rather, you need the full information of a coordinate system $(U,(x^1,\dots, x^n))$ and from this entire coordinate chart you can construct $n$-different vector fields $\frac{\partial}{\partial x^i}$.
Note very carefully that the definition of $\frac{\partial}{\partial x^1}$ depends on not just the coordinate function $x^1:U\to\Bbb{R}$, but on the remaining $(n-1)$ functions $x^2,\dots, x^{n}:U\to\Bbb{R}$ as well (but like I said above, the exterior derivative $dx^1\equiv d(x^1)$ makes sense regardless of $x^2,\dots, x^n$). So, perhaps it might be better to write the notation as $\left(\frac{\partial}{\partial x}\right)_1$ or $\left(\frac{\partial}{\partial x^1}\right)_{(x^1,\dots, x^n)}$ rather than simply $\frac{\partial}{\partial x^1}$ to indicate that it is the first vector field associated to the entire coordinate system $x=(x^1,\dots, x^n)$.
So, coming to the Schwarzschild vs Eddington-Finkelstein coordinates, you indeed have to be cautious of the coordinate $r$-vector field $\left(\frac{\partial}{\partial r}\right)_{(t,r,\theta,\phi)}$ associated to the Schwarzschild coordinates $(t,r,\theta,\phi)$ vs the coordinate $r$-vector field $\left(\frac{\partial}{\partial r}\right)_{(v,r\theta,\phi)}$ associated to the Eddington-Finkelstein coordinates $(v,r,\theta,\phi)$.
I want to stress this once again:
- the function $r$ makes sense all on its lonesome
- the exterior derivative $dr$ also makes sense all on its lonesome
- a vector field “$\partial_r=\frac{\partial}{\partial r}$” DOES NOT make sense all on its own. You MUST specify the remaining coordinates (because they specify the things you keep constant).
Here’s the same issue I emphasize in the thermodynamics context.
At the bare linear algebra level, you can already see this sort of single vs many dependence. Given a vector space $V$ and a non-zero vector $v$, I cannot speak of “the dual vector of $v$”. Rather, given an entire basis $\{e_1,\dots, e_n\}$ for $V$, I can consider its dual basis $\{\epsilon^1,\dots, \epsilon^n\}$. However, if I change my original basis say I replace $e_1$ by some other $w_1$, and consider the basis $\{w_1,e_2,\dots, e_n\}$, then I will get a corresponding dual basis $\{\xi^1,\dots,\xi^n\}$, and even though I only changed the first vector in my basis for $V$, there is no reason to assume the same will be true for the dual bases, i.e it need not be true that $\epsilon^2=\xi^2,\dots, \epsilon^n=\xi^n$.
Just for fun, let’s do some more computations. In the other answer, you see how the $r$-coordinate derivatives are related. Let me show you how the $t$ and $v$-coordinate derivatives are related (slightly different presentation): It’s essentially the chain rule (as you see in the other answer), but here I’ll present it in a slightly different way (but ‘really’ it’s the same idea, just phrased using duality). In general, for any vector field $X$ and any coordinate system $(U,x=(x^1,\dots, x^n))$ we can write $X=(dx^i)(X)\left(\frac{\partial}{\partial x^i}\right)_{(x^1,\dots, x^n)}$. So applied in our case,
\begin{align}
\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}&=
dv\left(\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}\right) \left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)}\\
&+
dr\left(\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}\right) \left(\frac{\partial}{\partial r}\right)_{(v,r,\theta,\phi)}\\
&+
d\theta\left(\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}\right) \left(\frac{\partial}{\partial \theta}\right)_{(v,r,\theta,\phi)}\\
&+
d\phi\left(\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}\right) \left(\frac{\partial}{\partial \phi}\right)_{(v,r,\theta,\phi)}
\end{align}
The second, third, and fourth terms are $0$, because we’re applying a $dx^j$ 1-form on the vector field $\left(\frac{\partial}{\partial x^j}\right)_{(x^1,\dots, x^n)}$ with $i\neq j$. For the first term, we note that $dv=dt+\frac{dr}{f(r)}$, so we get
\begin{align}
\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}&=
\left(dt+\frac{dr}{f(r)}\right)\left(\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}\right) \left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)}\\
&=\left(1+\frac{1}{f(r)}\cdot 0\right)\cdot \left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)}\\
&= \left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)}
\end{align}
Hence, $\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)} = \left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)} $. This is all the more a trippy result to think about ‘naively/intuitively’ especially when you consider that the level sets of the $t$ coordinate function are spacelike hypersurfaces, whereas the level sets of the $v$-coordinate function are null hypersurfaces. Anyway, what this shows is the striking difference between vector fields, 1-forms, exterior derivatives and gradient vector fields and their difference (and why you MUST be careful of their difference, and this isn’t just mathematical mumbo-jumbo for its own sake).
