It matters of two different concepts. Given a manifold, a vector is a geometric object associated to each point in the manifold. It can be decomposed into components with respect to a set of basis vectors.
$A = A^\mu \hat e_{(\mu)}$
where:
$\mu = 0, 1, 2, 3$
$A$ vector
$A^\mu$ contravariant components
$\hat e_{(\mu)}$ basis vectors
The geometric object is a reality independently of the coordinate system. A characterization is given by its square.
$A^2 = A \cdot A = A^\mu \hat e_{(\mu)} \cdot A^\nu \hat e_{(\nu)} = \hat e_{(\mu)} \cdot \hat e_{(\nu)} A^\mu A^\nu = g_{\mu\nu} A^\mu A^\nu$
where:
$\cdot$ scalar (dot) product
$g_{\mu\nu} = \hat e_{(\mu)} \cdot \hat e_{(\nu)}$ metric tensor
The square can also be written as
$A^2 = A_\mu A^\mu$
where:
$A_\mu = g_{\mu\nu} A^\nu$
As per above, we can define the dual vector.
$\tilde A = A_\mu \hat \theta^{(\mu)}$
where:
$\tilde A$ dual vector
$A_\mu$ covariant components
$\hat \theta^{(\mu)}$ basis dual vectors
By demanding
$\hat \theta^{(\mu)} (\hat e_{(\nu)}) = \delta^\mu_\nu$
where:
$\delta^\mu_\nu$ Kronecker delta
we can write the action of the dual vector on the vector as
$\tilde A (A) = A_\mu \hat \theta^{(\mu)} (A^\nu \hat e_{(\nu)}) = A_\mu A^\nu \hat \theta^{(\mu)} (\hat e_{(\nu)}) = A_\mu A^\nu \delta^\mu_\nu = A_\mu A^\mu$
Hence, a dual vector is a linear map from the vector space to the real numbers.
By defining the inverse metric tensor as
$g^{\mu\lambda} g_{\lambda\nu} = \delta^\mu_\nu$
where:
$g^{\mu\nu}$ inverse metric tensor
we have also
$A^\mu = g^{\mu\nu} A_\nu$
