I am having some trouble coming to terms with the notion of a derivative operator on a manifold. In Robert M. Wald's General Relativity, the definition in the textbook is given in terms of 5 stipulations:
Linearity: For all tensors $\mathit{A}, \mathit{B} \in \mathscr{T}(k,l)$ and $\alpha, \beta \in \mathbb{R},$ $$\nabla_c ( \alpha \mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} + \beta \mathit{B}^{a_1...c...a_k}_{b_1...c...b_l}) = \alpha \nabla_c \mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} + \beta \nabla_c \mathit{B}^{a_1...c...a_k}_{b_1...c...b_l} .$$
Leibnitz rule: For all $\mathit{A} \in \mathscr{T}(k,l),$ $\mathit{B} \in \mathscr{T}(k',l'),$ $$\nabla_e [\mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} \mathit{B}^{c_1...c...c_k'}_{d_1...c...d_l'}] = \nabla_e [A]B + A \nabla_e[B] .$$
Commutativity with contraction: For all tensors $\mathit{A} \in \mathscr{T}(k,l)$, $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) = \nabla_{d}\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}$.
Consistency with the notion of tangent vectors as directional derivatives on scalar fields: for all functions $f \in \mathscr{F}:M \rightarrow \mathbb{R}$ and all tangent vectors $t^a \in V_p$, it is required that $t(f)=t^a\nabla_a f.$ Note that this implies the action of a derivative operator will agree in the action on any scalar field.
(Optional) Torsion free: For all $f \in \mathcal{F}$ $$\nabla_a \nabla_b f = \nabla_b \nabla_a f.$$
Immediately thereafter, it is shown that derivative operators exist by considering the "ordinary derivative operator" $\partial_a$ in some coordinate system. However, I am not sure why ordinary derivatives will satisfy condition 4. That is, I am not sure why ordinary derivative operators will agree in their action scalar fields
For example, suppose our manifold is the plane $\mathbb{R}^2,$ with one coordinate system being the usual cartesian coordinate system, and the other coordinate system being the usual polar coordinate system.
Consider the function $f(x,y) = xy.$ Consider the tangent vector $\hat{y}$. The ordinary derivative operator we write $\frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y}.$ Then $t(f) = \hat{y} \cdot (y \hat{x} + x\hat{y}) = x.$
Now consider this in the polar coordinate system. $\tilde{f} (r, \theta) = r^2 \cos \theta \sin \theta.$ Consider the same tangent vector $\hat{y} = \sin\theta \hat{r} + \cos \theta \hat{ \theta}.$ Now consider the ordinary derivative operator $\frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \hat{\theta}.$ Note that this is different than what you would get if you wanted to consider what the gradient is in polar coordinates. Then $$t(\tilde{f}) = (\sin\theta \hat{r} + \cos \theta \hat{ \theta}) \cdot (2r \cos \theta \sin \theta \hat{r} + r^2(\cos^2 \theta - \sin^2 \theta) \hat{\theta}).$$ But note that $t(f) \neq t(\tilde{f}).$
So how do we have condition 4?