Your argument is incomplete. There are two problematic steps, which both contain errors. Firstly, it's not clear that $1_A$ should correspond to an isomorphism on the right hand side (and this is not generally true). Secondly, $h^A\cong F$ doesn't contradict anything.
Indeed, there are categories $\mathcal{C}$ for which $\mathcal{C}(-,A)$ can be made a covariant functor (for all $A$). In particular, any category with a functor $\bar{\cdot}:\mathcal{C}\to \mathcal{C}^{\text{op}}$ which is the identity on objects will do, since then we get a covariant functor from $X\mapsto \mathcal{C}(\bar{X},A)$. A classic example of this is given by groups (regarded as single object categories w/ object $*$), $g\mapsto g^{-1}$ defines a contravariant functor on a group $G$ regarded as a category since $(gh)^{-1}=h^{-1}g^{-1}$.
A nontrivial example of such a category is the category of inner product spaces with all linear maps between them. The adjoint of a linear map defines a contravariant functor which is the identity on objects.
When we define $F$ this way, the natural transformation $h^A\to F$ is going to be defined by
$$f\in h^V(W) = \mathcal{C}(V,W)\mapsto \bar{f} \in F(W) = \mathcal{C}(W,V).$$
To construct an $F$ for which this is not an isomorphism, we need a functor $\mathcal{C}\to\mathcal{C}^{\text{op}}$, which is the identity on objects, but is not fully faithful.
For this, we could take $G$ a group, and define $\bar{g} = 1$ for all $g$. This also defines a contravariant functor, though admittedly a rather boring one. Thus we see that the natural map $h^A\to F$ isn't necessarily an isomorphism.
What is true
You've asked why $\mathcal{C}(-,A)$ must be contravariant. The answer is sort of simple, it's just that it is contravariant. We don't define it to be contravariant, it just naturally has that structure. We can put other structures on the map from objects to sets to make it a covariant functor, as we've just seen, but it would still have the natural contravariant functor structure.
We can also make a more general statement. It is not always possible to put a covariant structure on the sets given by $\mathcal{C}(-,A)$.
Proof.
It suffices to give a counterexample. Consider the category with two objects, $0$, and $1$, and a single nonidentity map, $f$, between them. Call this category $\newcommand\2{\mathbf{2}}\2$.
Then $\2(0,0)=\{1_0\}$, but $\2(1,0)=\varnothing$, so if we had a covariant structure on $\2(-,0)$, we would get a map $f_*: \{1_0\}\to \varnothing$, which is impossible.
Best Answer
Note that $\mathcal A$ is really the same thing as $(\mathcal A^{\mathrm{op}})^{\mathrm{op}}$. So, if we apply the second definition to a functor $X : \mathcal A = (\mathcal A^{\mathrm{op}})^{\mathrm{op}} \to \mathsf{Set}$, we get that we call it representable if $$ X \cong \mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(-, A) \cong \mathrm{Hom}_{\mathcal A}(A, -). $$ However, I think this perspective is a lot less useful than the perspective in the comments, that these are really just two separate definitions.