My professor in his notes claims that a functor $G$ has a right adjoint iff the functor $Y \mapsto Hom(X, G(Y)$ is corepresentable, i.e for each $X$ there is an object $F(X)$ and a nutural by $X$ bijection
$Hom(F(X), Y) \cong Hom(X, G(Y))$.
Is that true? Of course, from corepresentability follows that $X \mapsto F(X)$ is indeed a functor (functoriality by $X$ plus Yoneda lemma gives us a unique arrow $F \phi : F(X) \to F(X')$ for each $\phi: X \to X'$).But I don't see how do we get functoriality by Y.
If it was true it would be enough to check if two given functors is adjoint pair to verify functoriality just by one argument. I never heard of it and always checked both arguments.
Yes, this is true. It is proven, for instance, in Theorem IV.1.2(ii)/Corollary IV.2.2 of Mac Lane's Categories for the Working Mathematician. In fact, one does not even need $G$ to be a functor: see Theorem 2 of Street's The core of adjoint functors.
However, note that this does not imply, when checking two functors are adjoint, that it is enough to check naturality in one of (or neither of) the arguments. The point of the theorem is that the action of $F$ on morphisms is uniquely determined by the hom set isomorphism, in such a way that the isomorphism is natural. If you start with a functor $F$ instead of just an assignment on objects, then you need to check naturality to ensure that the action of the functor $F$ on morphisms coincides with the canonical choice given by the hom-set isomorphism.
In other words, this theorem is useful when you want to construct an adjoint, rather than when you already have two functors and want to check that they are adjoint.