Here is the proof:
But I do not understand the following:
1-why the function needed to be bilinear to use the universal property?
2- what is he doing starting from the paragraph that starts with the statement "In the other direction …." ?
A difficulty in understanding the proof of distributivity of tensor products over direct sums for modules.
direct-summodulesrepresentation-theorytensor-products
Related Solutions
For the best argument, see the comment by Qiaochu Yuan.
Here is an alternative argument: Let $R$ be a comutative ring. I assume that you already know that the tensor product of $R$-modules commutes with direct sums (see below for the isomorphism). Now let $G$ be a group and let $A,B,C$ be representations of $G$ over $R$, i.e. $R$-modules with an action of $G$. Then the isomorphism of $R$-modules
$f : (A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C), ~ f(a \otimes b)=a \otimes (b,0),~ f(a \otimes c)=a \otimes (0,c)$
is also $G$-equivariant, because $f(g(a \otimes b))=f(ga \otimes gb)=ga \otimes (gb,0)=ga \otimes g(b,0) = g f(a \otimes b)$ and likewise $f(g(a \otimes c))=g f(a \otimes c)$.
Actually, $f$ commutes with the $G$-actions simply because $f$ is a natural isomorphism. Even more abstractly, such a morphism $f$ exists in any monoidal category with coproducts, which is here the category ${}^G \mathsf{Mod}(R)$ of representations of $G$ over $R$. Therefore, actually no computation is needed at all.
Here is a generalization: Let $\mathcal{C}$ be a monoidal category and let $G$ be an arbitrary small category. Then the category ${}^G \mathcal{C}$ of functors $G \to \mathcal{C}$ is again monoidal, the tensor product is defined objectwise. If $\mathcal{C}$ has direct sums which commute with $\otimes$, then the same is true for ${}^G \mathcal{C}$, for trivial reasons: For $A,B,C \in {}^G \mathcal{C}$ there is a canonical morphism $(A \otimes B) \oplus (A \otimes C) \to A \otimes (B \oplus C)$ in ${}^G \mathcal{C}$. That it is an isomorphism, may be checked objectwise, and thereby reduced to $\mathcal{C}$.
The property expressed in Theorem 8 is an adjunction of functors; precisely the functor $S\otimes_R-\colon R\text{-Mod}\to S\text{-Mod}$ (extension of scalars) is a left adjoint to the functor $\iota\colon S\text{-Mod}\to R\text{-Mod}$ (restriction of scalars).
In a very broad sense, $R\text{-Mod}$ is “reflected” in $S\text{-Mod}$ and conversely, but something is generally lost in these correspondences. An example is $R=\mathbb{Z}$ and $S=\mathbb{Q}$; if $N=\mathbb{Z}/2\mathbb{Z}$, then $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z}=0$ and the theorem basically says that there is no nontrivial homomorphism of $\mathbb{Z}/2\mathbb{Z}$ into $L$, where $L$ is any $\mathbb{Q}$-module.
It is a special case of the more general adjunction between tensor functors and Hom functors.
The second part in your question has nothing to do with Theorem 8. It just states that $$ (M\oplus M')\otimes_R N\cong (M\otimes_RN)\oplus(M'\otimes_R N) $$ (and the symmetric situation when the direct sum is on the opposite side).
Best Answer
The universal property of tensor product basically says that there is a one-to-one correspondence between $$\matrix{M\times N\to T&R\text{-bilinear}\\ \hline M\otimes N\to T& R\text{-linear}} $$ maps where $M, N, T$ are arbitrary $R$-modules.
That's why for finding $(M\oplus M')\otimes N\to (M\otimes N)\oplus (M'\otimes N)$ we actually only have to define a bilinear map $(M\oplus M')\times N\to(M\otimes N)\oplus(M'\otimes N)$.
Similarly, for the other direction, the two bilinear maps $M\times N\to(M\oplus M')\otimes N$ and $M'\times N\to (M\oplus M') \otimes N$ induce $R$-homomorphisms with domains $M\otimes N$ and $M'\otimes N$, which together induce a map from their direct sum, actually by the universal property of coproducts (and that one indeed doesn't involve bilinear maps).