Joint distribution of uniform variables $$X_{1}\sim \mathcal{Uniform}([0,2])$$
$$X_{2}\sim \mathcal{Uniform}([1,2])$$
$X_{1}$ and $X_{2}$ are independent random variables.
My question is whether joint distribution of those variables is:
$$f(x_{1},x_{2})=f_{1}(x_{1})\,f_2(x_{2})=1/2$$
... for all $(x_1, x_2) \in\ [0,2]{\times}[1,2]$?
If answer is incorrect, then how to approach this issue? 
If answer is correct, then I am thinking wheter exist more rigorous and formal approach to derive this?
 A: Short answer: Yes, you are correct.
But you asked for rigor... I'll try to make the rationale rigorous for your edification :P Hopefully I don't end up confusing you, but rather send you on an adventure to learn more formal probability theory.
Let $(\Omega, \mathscr{F}, P)$ be our underlying probability space (meaning all random variables we discuss here are assumed to be $\mathscr{F}$-measurable functions of $\omega \in \Omega$).
Consider the following random variable $X: \Omega \to \mathbb{R}^2$,
$$X = \begin{bmatrix}X_1 \\ X_2\end{bmatrix}$$
Notice that the components of $X$ are also random variables, $X_1: \Omega \to \mathbb{R}$ and $X_2: \Omega \to \mathbb{R}$.
Let the probability density function (PDF) of $X$ be called $f(x) = f(x_1,x_2)$, and let the PDFs of its components $X_1$ and $X_2$ be called $f_1(x_1)$ and $f_2(x_2)$ respectively. We define a conditional PDF as,
$$f_1(x_1 | x_2) := \frac{f(x_1, x_2)}{f_2(x_2)}$$
When we say "$X_1$ and $X_2$ are independent" we strictly mean,
$$f_1(x_1 | x_2) \equiv f_1(x_1)$$
So like you said,
$$f(x_1, x_2) = f_1(x_1)f_2(x_2)$$
If $f_1(x_1)$ is uniform on $[0,2]$ then,
$$f_1(x_1)
:= \begin{cases}
\frac{1}{2}, & x_1 \in [0,2] \\
0, & \text{else}
\end{cases}$$
and similarly, $f_2(x_2)$ uniform on $[1,2]$ means,
$$f_2(x_2)
:= \begin{cases}
1, & x_2 \in [1,2] \\
0, & \text{else}
\end{cases}$$
So we must have,
$$f(x_1,x_2) = f_1(x_1)f_2(x_2)
= \begin{cases}
(\frac{1}{2})(1), & x_1 \in [0,2]\ \cap\  x_2 \in [1,2] \\
(\frac{1}{2})(0), & x_1 \in [0,2]\ \cap\  x_2 \not\in [1,2] \\
(0)(1), & x_1 \not\in [0,2]\ \cap\  x_2 \in [1,2] \\
(0)(0), & x_1 \not\in [0,2]\ \cap\  x_2 \not\in [1,2] \\
\end{cases}$$
which can be expressed more simply as,
$$f(x_1,x_2)
= \begin{cases}
\frac{1}{2}, & x_1, x_2 \in [0,2]\times[1,2] \\
0, & \text{else}
\end{cases}$$
It is good to verify that your result is indeed a valid PDF,
$$\iint_{\mathbb{R}^2}f(x)dx = \int_1^2 \int_0^2\frac{1}{2}dx_1dx_2 = 1$$
