Associating to any vector space its $k$-linear dual and the resulting functor From my understanding, a contravariant functor of a category $\mathcal{C}$ can be defined using the notion of opposite category, $\mathcal{C}^{op}$. Then for two categories $\mathcal{C}$ and $\mathcal{D}$, the contravariant functor $\mathcal{C} \rightarrow \mathcal{D}$ is the covariant functor $\mathcal{C}^{op} \rightarrow \mathcal{D}$.
Now let $\mathcal{C}$ be the category of vector spaces $V$ over a field $k$. Take any vector space $V$ and associate to it its $k$-linear dual, $V^{*}$.
How can I see whether this correspondence results in a covariant or contravariant functor?
I think my confusion arises from the fact that both $\mathcal{C}$ and $\mathcal{C}^{op}$ consist of the same objects.
Note that, unfortunately, my understanding of Category Theory is basic.
 A: The main idea is that a contravariant functor reverses the arrows.
As to your concrete example: if $f:V\to W$ is a linear transformation, then the dual map $f^*:W^*\to V^*$.

Edit. (Some more details.) A covariant functor $F:\mathscr{C}\to \mathscr{D}$ is such that $F(gf)=F(g)F(f)$ for $f:X\to Y$ and $g:Y\to Z$.
A contravariant functor $F':\mathscr{C}\to \mathscr{D}$ is such that $F'(gf)=F'(f)F'(g)$.
In the dual category $\mathscr{C}^{\text{opp}}$, we have $$\text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Y):=\text{Hom}_{\mathscr{C}}(Y,X)$$
and composition is done 'the other way around':
$\text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Y)\times \text{Hom}_{\mathscr{C}^{\text{opp}}}(Y,Z)\to \text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Z),(f,g)\mapsto fg.$
Hence, we can reformulate by saying that $F'$ is a covariant functor $\mathscr{C}^{opp}\to \mathscr{D}$

As to the example of the dual vector space, we take 
$$f:V\to W,g:W\to Z,$$
then 
$$Ff:W^*\to V^*,Fg:Z^*\to W^*$$
and
$$F(gf):Z^*\to V^*,$$
so $F(gf)=F(f)F(g)$. This is how we see that $(-)^*$ is contravariant.
