Show map between tensor products is a $R-$algebra homomorphism

Prove that, if $A,B,C,D$ are algebras over a commutative ring R, and $f : A \rightarrow B, g : C \rightarrow D$ are homomorphisms of $R-$algebras, then $f \otimes g : A \otimes C \longrightarrow B \otimes D$ is a homomorphism of $R-$algebras.

I know I have to prove the map is a ring homomorphism and a $R-$module homomorphism. The "scalar" property seems easy. We take $r\in R, x\in A, y\in C$:

$$(f\otimes g)(r(x\otimes y))=(f\otimes g)(rx\otimes y)=f(rx)\otimes g(y)=rf(x)\otimes g(y)=r(f(x)\otimes g(y))$$

Then, for the product, take $x,z\in A, y, w\in C$: $$(f\otimes g)((x\otimes y)(z \otimes w)) = (f\otimes g)(xz \otimes yw)= \\=f(xz)\otimes g(yw)= f(x)f(z)\otimes g(y)g(w) =\\ =(f(x)\otimes g(y))(f(z)\otimes g(w))=(f\otimes g)(x\otimes y)(f\otimes g)(z \otimes w)$$

My problem is checking that, for $x,z\in A, y, w\in C$ we have: $$(f\otimes g)((x\otimes y)+(z\otimes w))= (f\otimes g)(x\otimes z)+(f\otimes g)(z\otimes w)$$

I must be missing something here about the addition of decomposable tensors... Can someone help with this?

Typically, $f\otimes g$ is defined using the universal property of the tensor product, as the unique $R$-module homomorphism $A\otimes C\to B\otimes D$ such that $(f\otimes g)(x\otimes y)=f(x)\otimes g(y)$ for all $x\in A$ and $y\in C$. So, by definition it is automatically an $R$-module homomorphism, and so you don't need to do any work to check that part. To put it another way, $f\otimes g$ is defined by setting $(f\otimes g)(x\otimes y)=f(x)\otimes g(y)$ and then extending to the rest of $A\otimes C$ such that it preserves addition (and the universal property guarantees that this extension is actually well-defined).