Isomorphism of bundles , Madsen's Calculus to Cohomology Let $\xi=(E,M,p)$ be a smooth bundle over $M$. Denote $\Omega^0(\xi)$ the smooth sections. 
We also define $\Omega^i(M):= \Omega^0(\wedge^i T^*M)$, differential $i$ forms, and $\Omega^0(M)$ is then smooth functions on $M$ to $\Bbb R$. 

Then in Madsen's book Calculus to Cohomology, page 171, it states 

$$\operatorname{Hom}_{\Omega^0(M)}(\Omega^0(\xi), \Omega^2(M) \otimes_{\Omega^0(M)} \Omega^0(\xi)) \cong  \operatorname{Hom}_{\Omega^0(M)}(\Omega^0(\xi), \Omega^0(\xi)) \otimes _{\Omega^0(M)} \Omega^2(M)$$


I don't understand why this is true. In fact, if we relabel everything, 
$$\operatorname{Hom}_R(A,B \otimes_RS) \cong \operatorname{Hom}_R(A,B) \otimes_R S$$ 
is what we have to show. But there is only a map clear from one direction, from RHS to LHS, 
$$ (f \otimes s ) (a):= f(a) \otimes s $$ 

How does one prove the isomorphism? 
 A: Well, the book doesn't just state this fact; it cites Theorem 16.13 from earlier in the book as justification.  That theorem says that the operation of tensoring and Homming sections of vector bundles over the ring $\Omega^0(M)$ corresponds to the tensor and Hom operations on the vector bundles themselves.  In particular, there are natural isomorphisms $$\operatorname{Hom}_{\Omega^0(M)}(\Omega^0(\xi), \Omega^2(M) \otimes_{\Omega^0(M)} \Omega^0(\xi))\cong \Omega^0(\operatorname{Hom}(\xi,\wedge^2T^*M\otimes \xi))$$ and $$\operatorname{Hom}_{\Omega^0(M)}(\Omega^0(\xi), \Omega^0(\xi)) \otimes _{\Omega^0(M)} \Omega^2(M)\cong \Omega^0(\operatorname{Hom}(\xi,\xi)\otimes\wedge^2T^*M).$$
The desired isomorphism then follows from the isomorphism of vector bundles $\operatorname{Hom}(\xi,\wedge^2T^*M\otimes \xi)\cong \operatorname{Hom}(\xi,\xi)\otimes\wedge^2T^*M$, which on fibers is just the natural isomorphism of vector spaces $\operatorname{Hom}(V,W\otimes V)\cong \operatorname{Hom}(V,V)\otimes W$.
