I'm having some trouble with the following problem.
Let $x, y$ be two commuting variables, and let $\mathbb K$ a field. We define the following function $$\mathbb K[x] \times \mathbb K[y] \rightarrow \mathbb K[x,y], \, (P(x),Q(y))\mapsto P(x)Q(y).$$
I have two questions.
1.) I don't see why this function is bilinear.
2.) Why can we say that our function induces an isomorphic $\mathbb K[x] \otimes \mathbb K[y] \rightarrow \mathbb K[x,y]?$
(For question 2., can we use the universal property of the tensor product?)
Your question makes little sense. You also need to specify what kind of isomorphism. Of vector spaces over $\mathbb{K}$? As $\mathbb{K}$-algebras? As $\mathbb{Z}$-algebras (= rings)?
You can show that $$\mathbb{K}[x] \otimes \mathbb{K}[y]\cong \mathbb{K}[x,y]$$ as $\mathbb{K}$-algebras.
I'll give a brief sketch. You can use the comments to ask for clarification if necessary.
Define the mapping $$\mathbb{K}[x] \times \mathbb{K}[y]\to \mathbb{K}[x,y]: (P(x), Q(y)) \mapsto P(x)Q(y)$$
This is a bilinear map by distributivity in $\mathbb{K}[x,y]$. The universal property of the tensor product gives a unique linear map $$\mathbb{K}[x] \otimes \mathbb{K}[y] \to \mathbb{K}[x,y]: P(x) \otimes Q(y) \mapsto P(x)Q(y)$$
You have to show this map is a bijection and you will be done. Surjectivity ought to be clear, but you need to work a bit to show injectivity.
One possible approach (see the top comment for another approach): You can try to define an explicit inverse. For example, define $$\mathbb{K}[x,y] \to \mathbb{K}[x] \otimes \mathbb{K}[y]$$ by $x \mapsto x \otimes 1$ and $y \mapsto 1 \otimes y$ and extend multiplicatively and linearly.