Why is $ \hbox{Ext}_R^* (M,M) = H^*(\hbox{Hom}_R^*(P^*,P^*))$? Let me first fix some notation and conventions. 
Let $ R$ be a ring and $ M$ a left $R$-module. Given chain complexes $P^*$ and $Q^*$ in $R$-mod, define $ \hbox{Hom}^*_R(P^*,Q^*)$ to be the graded space such that $ \hbox{Hom}_R ^i (P^*, Q^*) = \bigoplus_n \hbox{Hom}_R(P^n, Q^{n-i})$. $\hbox{Hom}^*_R (P^*,Q^*)$ has a differential with degree $+1$. Now let $ P^*$ be a projective resolution of $M$. 

Why is $ \hbox{Ext}^*_R (M,M)= H^* ( \hbox{Hom}_R^* (P^*, P^*))$?

I've seen this result stated in a few places but I dont see why it's true. I think  I could believe that $ \hbox{Ext}^*_R (M,M) = H^* ( \hbox{Hom}_R^*(P^*, M))$. But why is there a projective resolution in the second slot of the Total Hom and not an injective resolution?
 A: They key here is that the differential of $\hom^\bullet(P_\bullet, Q_\bullet)$ is designed so that
$$H^i(\hom^\bullet(P_\bullet, Q_\bullet)) = \hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i])$$
where $\mathsf{K}(R)$ is the homotopy category (you might want $i$ instead of $-i$, it depends on your indexing conventions and I can't remember what you should get in this case).
If $P_\bullet$ is a projective resolution then it is colocal with respect to acyclic complexes.  This is what Daniel Murfet calls a hoprojective resolution (google that term to find his notes on derived categories) and it has the property that
$$\hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, Q_\bullet[-i])$$
where $\mathsf{D}(R)$ is the derived category.  Now if $Q_\bullet = P_\bullet$ then it is quasi-isomorphic to $M$ as a complex ($M$ as a complex has an $M$ in the zeroth position and has $0$'s elsewhere).  Quasi-isomorphisms in $\mathsf{D}(R)$ are actually isomorphisms so
$$H^i(\hom^\bullet(P_\bullet, P_\bullet)) = \hom_{\mathsf{D}(R)}(P_\bullet, P_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, M[-i]) = H^i(\hom^\bullet(P_\bullet, M))$$
Now just check that $\hom^\bullet(P_\bullet, M)$ when $M$ is considered a complex is the same as $\hom(P_\bullet, M)$ when $M$ is merely a module, and the homology of that last thing becomes $\mathrm{Ext}^i(M, M)$.
Edit: Just fyi, my experience is with dg-algebras in characteristic $2$, so forgive me if I've left out any necessary $\pm1$'s, because they don't matter to me!
A: I want to justify the answer in user52045's comment by a reference: Brown, Cohomology of Groups, Theorem I.8 (8.5) states: 

If $f: C \to D$ is a quasi-isomorphism between arbitrary complexes and $P$ is a non-negative complex of projectives (over an ring $R$), then $Hom^\ast_R(P,f): Hom^\ast_R(P,C)\to Hom^\ast_R(P,D)$ is a quasi-isomorphism. 

Now let $M[0]$ be the complex having $M$ in degree 0 and zero elsewhere. Thus a projective resolution $P \to M$ is a quasi-isomorphism $f:P \to M[0]$ and by the theorem above: 
$$H^\ast Hom^\ast_R(P,P) \cong H^\ast Hom^\ast_R(P,M[0]) = H^\ast Hom_R(P,M)=Ext^\ast_R(M,M)$$
