I'm somewhat confused on the following calculation in the proof of Artin-Wedderburn: let $R$ be a semisimple ring such that (since $R$ is f.g. over itself) we have $R \simeq \oplus_{i = 1}^r S_i^{n_i}$. Now,
$$ R^{op} \stackrel{(1)}\simeq End_R(R) \simeq End_R(\oplus_{i = 1}^r S_i^{n_i}) \stackrel{(2)}\simeq \bigoplus_{i = i}^rEnd_R(S_i)^{n_i \times n_i} $$
Now, I know that $(1)$ is a ring isomorphism between $R^{op}$ and the left $R$-module morphisms from $R$ to istelf, but why is $(2)$ a ring isomorphism? I know we have $Z(R)$-module isomorphism between $Hom_R(M \oplus N,P)$ and $Hom_R(M,P) \oplus Hom_R(N,P)$, so $(2)$ could be stated as a module isomorphism, but however $(1)$ is a ring homomorphism so it would not make sense in this context.
You just need to look deeper at the theorem you mentioned.
The simplest example of how it works looks like this:
$End(M\times N)\cong \begin{bmatrix}Hom(M, M)& Hom(N,M)\\Hom(M,N)& Hom (N,N)\end{bmatrix}$ as rings where the multiplication is matrix multiplication. Take a look at how the elements from the various entries compose with each other and generalize it to your situation.
The final piece of the puzzle is to notice that when $M$ and $N$ are different Wedderburn components of $R$, there is no nonzero homomorphism between them. In our toy example above, if $Hom(M,N)=Hom(N,M)=\{0\}$, then you're looking at the diagonal matrix ring
$\begin{bmatrix}Hom(M, M)& 0\\0& Hom (N,N)\end{bmatrix}\cong Hom(M,M)\times Hom(N,N)=End(M)\times End(N)$, all as ring homomorphisms.