differential geometry - Is partial derivative a vector or dual vector?

The textbook(Introduction to the Classical Theory of Particles and Fields, by Boris Kosyakov) defines a hypersurface by $$F(x)~=~c,$$ where $F\in C^\infty[\mathbb M_4,\mathbb R]$. Differentiating gives $$dF~=~(\partial_\mu F)dx^\mu~=~0.$$ The text then says $dx^\mu$ is a covector and $\partial_\mu F$ a vector. I learnt from another book that $dx^\mu$ are 4 dual vectors(in Minkowski space), $\mu$ indexes dual vector themselves, not components of a single dual vector. So I think $\partial_\mu F$ should also be 4 vectors, each being the directional derivative along a coordinate axis. But this book later states that $(\partial_\mu F)dx^\mu=0$ describes a hyperplane $\Sigma$ with normal $\partial_\mu F$ spanned by vectors $dx^\mu$, and calls $\Sigma$ a tangent plane (page 33-34). This time, it seems to treat $\partial_\mu F$ as a single vector and $dx^\mu$ as vectors. But I think $dx^\mu$ should span a cotangent space.

I need some help to clarify these things.

[edit by Ben Crowell] The following appears to be the text the question refers to, from Appendix A (which Amazon let me see through its peephole):

Elie Cartan proposed to use differential coordinates $dx^i$ as a convenient basis of 1-forms. The differentials $dx^i$ transform like covectors [...] Furthermore, when used in the directional derivative $dx^i \partial F/\partial x^i$, $dx^i$ may be viewed as a linear functional which takes real values on vectors $\partial F/\partial x^i$. The line elements $dx^i$ are called [...] 1-forms.