What explicitly is the "adjunction" isomorphism $Hom_C(A,C)\to Hom_B(A,Hom_C(B,C))$? Suppose $B$ and $C$ are commutative rings, $A$ a $B$-algebra, and $B$ is a $C$-module. What exactly is the "adjunction" isomorphism 
$$Hom_C(A,C)\to Hom_B(A,Hom_C(B,C))?$$

Given $A\to C$, it needs to be composed with a morphism $C\to Hom_C(B,C)$. I know $Hom_C(C,B)\simeq B$, but that doesn't seem to work here. 

Here also $A$ is a finitely generated projective $B$-module, and $B$ is a finitely generated projective $C$-module.
I believe the adjunction in question is likely the classic tensor hom adjunction $\text{Hom}_C(A \otimes_R B, X) \cong \text{Hom}_R(A, \text{Hom}_C(B, X))$. Where all the objects in question have the appropriate structures such that this makes sense (X is a module over C, A is a module over R, B is a bi-module over both R and C, and R, C are just rings). 

The statement you are trying to prove then follows from the observation that if $A$ is a $B$ algebra, then $A \otimes_B B \cong A$ as $C$-modules, taking $X = C, R = B$ above. This is a very simple thing to show if you understand the tensor product.  

Edit: the explicit map takes a $\phi \in \text{Hom}_C(A \otimes_B B, C)$ and sends this to the map $\psi \in \text{Hom}_B(A, \text{Hom}_C(B, C))$ where $\psi$ sends $a \to \phi_a$ where $\phi_a(b) = \phi(a \otimes b)$. That this mapping is $B$-linear and an isomorphism is once again not hard.
The actual isomorphism is (in category theory):

hom(AxB,C) = hom(A,hom(B,C))


It is normally known as "currying". (the hom's are in Set, not in C or B)

I'm a little confused why they call it "adjunction", since that means slightly different thing, but the word doesnt seem to explain your isomorphism.