Why isn't there a contravariant derivative? (Or why are all derivatives covariant?)

The "covariant" in "covariant derivative" should really be "invariant". It is a misnomer, but we are stuck with it. It is not the same "covariant" as that of a "covariant vector", and therefore, there is no "contravariant derivative". Armed with this, Wikipedia should fill in the rest for you :)

Let $S = C^\infty(X, \mathbb{R})$ be the $\mathbb{R}$-algebra of all smooth functions on the manifold X. There is a purely algebraic notion of Kähler differentials: there is an $S$-module $\Omega_{S/\mathbb{R}}$ presented by generators and relations as

• For every $f \in S$, add a generator $\mathrm{d}f$
• For every $r \in \mathbb{R}$, add the relation $\mathrm{d}r = 0$
• For every $f,g \in S$, add the relation $\mathrm{d}(f+g) = \mathrm{d}f + \mathrm{d}g$
• For every $f,g \in S$, add the relation $\mathrm{d}(fg) = f \mathrm{d}g + g \mathrm{d}f$

There are a few other ways to define it, but it's clear that $\Omega_{S / \mathbb{R}}$ is the "universal" way to differentiate things in $S$, if we presume that differentiation must be $\mathbb{R}$-linear and satisfy the Leibniz rule.

There is an obvious map $\Omega_{S / \mathbb{R}}$ to the global sections of $T^*X$, sending $\mathrm{d}f$ to $\mathrm{d}f$.

However, the Kähler differentials seem too big. However, for each point $P$, I will define an "evaluation map". Let $\Omega_{S/\mathbb{R}, P}$ be the $\mathbb{R}$-vector space you get by adding the further relations $f \mathrm{d}g = f(P) \mathrm{d}g$ for every pair of functions $f,g \in S$, and write $\mathrm{d}f|_P$ for the image of $\mathrm{d}f$ in this space.

Then, following the idea of this Mathoverflow answer, we can quickly show that $\Omega_{S / \mathbb{R}, x_0}$ is precisely the cotangent space at $x_0$. For any smooth $f$, we take the differential of a Taylor polynomial and get $$ \mathrm{d}f(x)|_{x_0} = f'(x_0) \mathrm{d} x|_{x_0}$$ (where $f'(x_0)$ is the linear functional $v \mapsto \nabla_v f(x_0)$) at which point it's clear that $\Omega_{S / \mathbb{R}, x_0} \cong T^*_{x_0} X$.

It's not hard to pass from $\mathbb{R}^n$ to manifolds. Consequently, we conclude that the exterior derivative is the "universal" way to differentiate smooth scalar fields in such a way that the derivative is completely determined by its "values" at points.

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Reading more of the answers, I think they claim further that $\Omega_{S / \mathbb{R}}^{**}$ is isomorphic to the global sections of the cotangent bundle as $S$-modules (and that this implies the global sections have what is maybe a nicer but less general universal property) but I haven't followed the argument.

The reason why the covariant derivative makes a $(p,q+1)$-type tensor field out of a $(p,q)$ type tensor field is because for a tensor field $T$, $\nabla T$ is defined as $$ \nabla T(X,\text{filled arguments})=\nabla_XT(\text{filled arguments}), $$ and $\nabla_XT$ is $C^\infty(M)$-linear in $X$, so this relation defines a covariant tensor field - one that acts on vector fields. But why does it need be so?

The geometric answer is that a covariant derivative is essentially a representation for a Koszul or principal connection, a device that allows for parallel transport of bundle data along curves. The reason it takes in vectors is because vectors are intrinsically tied to curves on your manifold. If your covariant derivative took in 1-forms as the directional argument instead of vectors, it would not represent a connection, because there is no way to canonically tie together curves and 1-forms without a tool like a metric tensor or a symplectic form.