Calculate the tensor product of two vectors

Let $\{e_1, e_2\}$ and $\{f_1, f_2, f_3\}$ the canonical ordered bases of $\mathbb{R}^2$ and $\mathbb{R}^3$ respectively. Find the coordinates of $x \otimes y$ with respect to the basis $\{e_i\otimes f_j\}$ of $\mathbb{R}^2\otimes\mathbb{R}^3$ where $x = (1, 1)$ and $y = (1, -2, 1)$.

Since I have that $(1, 1) = e_1 + e_2$ and $(1, -2, 2) = f_1 - 2f_2 + f_3$, then $x \otimes y = (1, 1) \otimes (1, -2, 1) = (e_1 + e_2)\otimes (f_1 - 2f_2 + f_3) = \dots ?$ How can I can compute this last tensor product? Thanks. Also any further reading recomendation about examples of simple tensor product product calculations and not only the abstract background would be appreciated ;)

Tensor product distributes over addition. That is, $$(a + b)\otimes c = a\otimes c + b\otimes c\quad \text{and}\quad a\otimes (b + c) = a\otimes b + a\otimes c.$$ Furthermore, for any real scalar $\lambda$, we have $$(\lambda a)\otimes b = a\otimes (\lambda b) = \lambda(a\otimes b).$$ These rules will allow you to write $x\otimes y$ as a linear combination of $e_i\otimes f_j$.
You can consider this type of calculation in a more general setting. If $x \in \mathbb{R}^m$ and $y \in \mathbb{R}^n$, their tensor product $x\otimes y$ is sometimes called their outer product. If $e_i\otimes f_j$ is the basis for $\mathbb{R^m}\otimes\mathbb{R}^n$ obtained from the standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$, then we have the following expression for the outer product:
$$x\otimes y = \sum_{i=1}^m\sum_{j=1}^na_{ij}e_i\otimes f_j$$
where $a_{ij}$ is the $(i, j)^{\text{th}}$ entry of the $m\times n$ matrix $xy^T$.
I suggest you simplify using the rules I gave at the beginning. Once you have the solution using that method, compute the matrix $xy^T$ and check that you get the same coefficients.
The tensor product behaves like multiplication in that it is distributive over addition. Additionally, scalars can be pulled to the front from either factor. These properties can be summed up as saying $(x,y)\mapsto x\otimes y$ is bilinear. We have $$(e_1+e_2)\otimes (f_1-2f_2+f_3)=e_1 \otimes f_1-2e_1 \otimes f_2+e_1 \otimes f_3+ e_2 \otimes f_1-2e_2 \otimes f_2+e_2 \otimes f_3$$