When we define the connection, we have to specify the connection coefficients for generalizing the directional derivative for a general $(p,q)$ tensor. However, when defining a derivative for antisymmetric tensors, we can directly define an exterior derivative without any consideration of parallel transport or how we subtract objects existing in two different tangent spaces.
Why is this?
Given a vector bundle $E\to X$ and a connection $\nabla$ on $E$, the covariant derivative is an operator $$\nabla: \Gamma(E)\to \Omega^1(X)\otimes \Gamma(E).$$ This operator is the first one of a sequence of operators $$d^\nabla:\Omega^k(X)\otimes \Gamma(E)\to \Omega^{k+1}(X)\otimes \Gamma(E)$$ for a definition see my answer here.
Now, let $E = X\times \mathbb R \to X$, the trivial line bundle of $X$, and choose as $\nabla$ the trivial (flat) connection (so the coefficient(s) will be zero in the natural trivialization of $E$), then $d^\nabla = d$ the exterior differential.