This is similar to what Serre wrote in his book on linear representations of finite groups by Serre:
There is also the tensor product which has the properties of "multiplication". Let $V_1$ and $V_2$ be two vector spaces. The tensor product $\otimes: V_1 \times V_2 \rightarrow V_1 \otimes V_2$ constructs a new vector space $V_1 \otimes V_2$ with the map $(\xi_1,\xi_2) \rightarrow \xi_1 \cdot \xi_2$ which can be defined as the set of formal linear combinations $\xi_1 \otimes \xi_2$ subject to the conditions:
$(i)$ $\xi_1 \cdot \xi_2$ is linear in both variables $\xi_1$ and $\xi_2$.
$(ii)$ If $\lbrace e_{i_1} \rbrace$ and $\lbrace e_{i_2} \rbrace$ is a basis for $V_1$ and $V_2$ respectively then $e_{i_1} \otimes e_{i_2}$ is a basis for $V_1 \otimes V_2$.
We can show that this space exits and is unique. By condition $(ii)$ we have:
$$
\dim(V_1 \otimes V_2) = \dim(V_1)\cdot \dim(V_2)
$$Now let $\rho^1\colon G \rightarrow GL(V_1)$ and $\rho^2\colon G \rightarrow GL(V_2)$ be two linear representations of a group $G$. For $s \in G$, define an element $\rho_s$ of $GL(V_1 \otimes V_2)$ by the condition:
$$
\rho_s(\xi_1 \cdot \xi_2) = \rho_s^1(\xi_1)\cdot \rho_s^2(\xi_2) \quad \text{for } \xi_1 \in V,\; \xi_2 \in V_2.
$$We write:
$$
\rho_s = \rho_s^1 \otimes \rho_s^2
$$The $\rho_s$ above defines a linear representation of $G$ in $V_1 \otimes V_2$ which is called the tensor product of the given representations.
Defining the above using matrix notation:
let $\lbrace e_{i_1} \rbrace$ be a basis for $V_1$ and let $r_{i_1j_1}(s)$ be the matrix of $\rho_s^1$ with respect to this basis, define $\lbrace e_{i_2} \rbrace$ and $r_{i_2j_2}(s)$ in a similar manner. Now we have:
$$
\rho_s^1(e_{i_1}) = \sum_{i_1} r_{i_1j_1}(s) \cdot e_{i_1}, \qquad \rho_s^2(e_{i_2}) = \sum_{i_2} r_{i_2j_2}(s) \cdot e_{i_2}
$$which implies:
$$
\rho_s(e_{i_1}\cdot e_{i_2}) = \sum_{i_1,i_2} r_{i_1j_1}(s) \cdot r_{i_2j_2}(s)\cdot e_{i_1} \cdot e_{i_2}
$$Consequently the matrix of $\rho_s$ is $(r_{i_1j_1}(s) \cdot r_{i_2j_2}(s))$ which is the tensor product of the matrices $\rho_s^1$ and $\rho_s^2$.
We must add, the tensor product of two irreducible representations is not in general irreducible. It decomposes into a direct sum of irreducible representations which can be determined by means of character theory, which we shall discuss in the next chapter.
My question therefore is, can someone construct a concrete example to make me understand this
Best Answer
Here's a rather simple but still informative example.
Let $\rho: \langle g: g^4=e \rangle \to GL(\mathbb{R}^2)$ be the representation of the cyclic group with four elements (i.e. $\mathbb{Z}/ 4\mathbb{Z}$) into $\mathbb{R}^2$ such that $$ \rho_g = \begin{bmatrix} 0 & -1 \\ 1 &0 \end{bmatrix} = M $$ and $\rho_{g^k}=M^k$. Note that there are no non-trivial $g$-invariant subspaces (one can verify this by showing $M$ does not have 1 as an eigenvalue) and so $\rho$ is irreducible.
However, let's look at what happens when we tensor $\rho$ with itself to get a representation of $\mathbb{Z}/ 4\mathbb{Z}$ into $\mathbb{R}^2 \otimes\mathbb{R}^2$. Letting $\rho' = \rho \otimes \rho$ we get that for $v,w\in \mathbb{R}^2$: $$ \rho'_g(v\otimes w) = \rho(v) \otimes \rho(w) $$ So your intuition should be that $\rho'$ mimics $\rho$ on each "copy" of $\mathbb{R}^2$. This is made clearer by actually computing the matrix of $\rho'_g$.
As you wrote, if $\{e_1,e_2 \}$ is the standard basis of $\mathbb{R}^2$ then the set $\{e_1\otimes e_1, e_1\otimes e_2, e_2\otimes e_1,e_2\otimes e_2\}$ forms a basis of $\mathbb{R}^2 \otimes \mathbb{R}^2$. Observe: $$ \rho'_g(e_1\otimes e_1) = Me_1 \otimes Me_1 = e_2 \otimes e_2 $$ $$ \rho'_g(e_1\otimes e_2) = Me_1 \otimes Me_2 = e_2 \otimes -e_1 = -e_2 \otimes e_1 $$ $$ \rho'_g(e_2\otimes e_1) = Me_2 \otimes Me_1 = -e_1 \otimes e_2 $$ $$ \rho'_g(e_2\otimes e_2) = Me_2 \otimes Me_2 = -e_1 \otimes -e_1 = e_1 \otimes e_1 $$ So the matrix of $\rho'_g$ is: $$ \rho'_g = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}. $$ This actually shows us something mildly interesting. While the original $\rho$ was an irreducible representation our new $\rho'$ is not since the subspace of $\mathbb{R}^2\otimes \mathbb{R}^2$ spanned by $\{e_1\otimes e_1, e_2\otimes e_2 \}$ is a proper invariant subspace. This confirms the statement at the end of your post.
Remark: I'm not sure if you know this but $\mathbb{R}^2\otimes \mathbb{R}^2 \cong \mathbb{R}^4$ as both are four dimensionals vector spaces over $\mathbb{R}$. So there really is nothing particularly fancy going on here, but I tried to keep everything in terms of tensors to help you see how things would work in a more general case.