Let $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ be (covariant) functors. Then $G$ is a left adjoint to $F$ if there exists a natural isomorphism
$$
\hom_{\mathcal{C}}(G(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F(Y)).
$$
For contravariant functors the same definition applies once we make them covariant. There are two ways for doing so, let's examine them.
So, let $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ be contravariant functors. We can get back to the previous situation by considering
$$\def\op{^{\mathrm{op}}}
F'\colon\mathcal{C}\to\mathcal{D}\op,\qquad
G'\colon\mathcal{D}\op\to\mathcal{C}
$$
and being $G'$ a left adjoint to $F'$ means there is a natural isomorphism
$$
\hom_{\mathcal{C}}(G'(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}\op}(X,F'(Y)).
$$
That's the same as saying there is a natural isomorphism
$$
\hom_{\mathcal{C}}(G(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(F(Y),X)
$$
(the functors are, in this case, adjoint on the left).
On the contrary, we could consider
$$
F''\colon\mathcal{C}\op\to\mathcal{D},\qquad
G''\colon\mathcal{D}\to\mathcal{C}\op
$$
and, being $G''$ a left adjoint to $F''$ means there is a natural isomorphism
$$
\hom_{\mathcal{C}\op}(G''(X),Y)
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F''(Y))
$$
which is the same as saying there is a natural isomorphism
$$
\hom_{\mathcal{C}}(Y,G(X))
\xrightarrow{\sim}
\hom_{\mathcal{D}}(X,F(Y))
$$
(the functors are adjoint to the right).
You can notice that, in case the categories are abelian, contravariant functors which are adjoint on the right are left exact; they are right exact in case they are adjoint on the left. I use the usual convention that the contravariant functor $F$ is called left exact if it transforms the exact sequence $0\to A\to B\to C\to 0$ into the exact sequence $0\to F(C)\to F(B)\to F(A)$.
There is no way to sensibly pair up $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$ if one is covariant and the other one is contravariant, because making them both covariant messes up either the domain or the codomain.
You may try all positions of $F$ and $G$ in the hom sets, but you'll not be able to compose maps in order to define “being natural”, whereas this is obviously possible for pairs of covariant or contravariant functors.
Best Answer
Note that $f: c \rightarrow c'$ is an arrow in $C$ if and only if $f^{\text{op}}: c' \rightarrow c$ is an arrow in $C^{\text{op}}$, so defining $F$ on arrows in $C$ is equivalent to defining $F$ on arrows in $C^{\text{op}}$.