This can be shown using the concept of oscillation.
For a bounded $f$, the oscillation of $f$ over set $A$ (which is not a single point) is given by
$$w(A) = \sup_{A} f - \inf_{A} f$$
For a single point $x$, the oscillation is defined as
$$ w(x) = \inf_{J} w(J)$$
where $J$ ranges over bounded intervals containing $x$.
Note that if $x \in I$, then $w(x) \le w(I)$.
Now if $f$ is Riemann integrable over $[a,b]$, then we can show that given any $n \gt 0$, there is a sub-interval $I_{n}$ of $[a,b]$ such that $w(I_n) \le \frac{1}{n}$.
This is because, if every subinterval $I$ of $[a,b]$ had $w(I) \gt \frac{1}{n}$, then for every partition of $[a,b]$ the difference between the upper and lower sums would be at least $\frac{b-a}{n}$ and as a consequence, $f$ would not be integrable.
Now pick $I_{n+1} \subset I_{n}$ such that $w(I_{n+1}) \le \frac{1}{n+1}$.
By completeness there is a point $c$ such that $c \in I_n$.
Thus $w(c) \le w(I_n) \le \frac{1}{n}$ for all $n$. Hence $w(c) = 0$.
Now it can be show that $f$ is continuous at point $x$ if and only if $w(x) = 0$.
Note: This is basically a simplification of one proof of the Riemann Lebesgue theorem of continuity almost everywhere.
What you write is more or less true. First note that it is not enough to assume that $f(x,t)$ is differentiable at $x_0$ (for (almost) all $t$), we need that there is a neighorhood $(x_0 - \varepsilon, x_0 + \varepsilon) =:I$ such that $f(\cdot, t)$ is differentiable in $x$ for all $x \in I$ and (almost) all $t \in A$, where the "exceptional null set" must not depend on $x$.
The assumption $|\partial f/\partial x (x,t)| \leq g(x)$ only makes sense in this setting, otherwise we could only require $x_0$ instead of $x$ and the proof also needs this assumption (see below).
Now, let $h_n \to 0$ with $h_n \neq 0$ for all $n$ and w.l.o.g. with $x_0 + h_n \in I$ for all $n$. Let
$$
F(x) := \int_A f(x,t) \, dt \text{ for } x \in I,\\
G(x) := \int_A \frac{\partial f}{\partial x}(x,t) \, dt \text{ for } x \in I.
$$
Here, the integral defining $F$ is interpreted as an improper Riemann integral (it is possible that this integral does not exist as a Lebesgue integral), but the integral defining $G$ is interpreted as a Lebesgue integral. If $t \mapsto \frac{\partial f}{\partial x}$ is Riemann integrable over every compact subinterval of $A$ (for example, if it is continuous in $t$), then it can also be interpreted as an improper Riemann integral, at least if we assume that the dominating function $g$ is (improper) Riemann integrable.
Observe that the assumptions guarantee that $F,G$ are well-defined. Measurability of $t \mapsto \frac{\partial f}{\partial x}(x,t)$ for each $x \in I$ is implied by the pointwise convergence $$\frac{\partial f}{\partial x}(x,t) = \lim_n \frac{f(x+h_n, t) - f(x,t)}{h_n},
$$
which will also be employed below.
We now have
\begin{eqnarray*}
\left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| & = & \left|\int_{A}\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\,{\rm d}t\right|\\
& \leq & \int_{A}\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\right|\,{\rm d}t.
\end{eqnarray*}
Note that the integrands are measurable and that for (almost) all $t \in A$, the mean value theorem yields some $\xi_{n,t} \in I$ (even between $x_0$ and $x_0 +h_n$) such that
$$
\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}\right|=\left|\frac{\partial f}{\partial x}\left(\xi_{n,t},t\right)\right|\leq g\left(t\right).
$$
This shows that the integrands in the integrals are Lebesgue-integrable and dominated by an integrable function. As the integrand in the last integral converges pointwise to $0$ for $n \to \infty$, the dominated convergence theorem yields
$$
\left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| \to 0,
$$
so that $F$ is differentiable (in $x_0$) with the expected derivative.
Best Answer
1 Does the derivative unbounded on $[−1,1]$ because of the zero value that $x$ may take in this interval and because of the angle $\frac{1}{x^2}$ in equation (1) and the term $x^{−3}$ in equation (1) also?
Yes, the the derivative is unbonded on $[−1,1]$. The term $2x \sin\frac{1}{x^2}$ is bounded on $(0,1]$ but if we evaluate the term $x^2(−2x^{−3})\cos\frac{1}{x^2}=−\frac{2}{x}\cos\frac{1}{x^2}$ in $a_k=\frac{1}{\sqrt{2 \pi k}}, k\in \Bbb N$ we have $−\frac{2}{a_k}\cos\frac{1}{a_k^2}=-2\sqrt{2 \pi k}$ so is not bounded.
2 Why we confined ourselves to the interval $[−1,1]$? Is there is a specific reason for that?
I think is because in every closed interval that contatins $0$ the derivate of $f$ is unbounded, so we can not use Riemann integration.
3 Does the Lebesgue integration guarantee for us that we can always integrate after differentiation? If so, how? which theorem guarantees this?
No. In fact the function $f'$ is not Lebesgue integrable. There are two conventions for Lebesgue integrability:
Unfortunately the function fails both conditions. Let $A_n= \{a_{k,n}\} ^{4n}_{k=0}$ where $a_{k,n}= \sqrt \frac{2}{ \pi(4n+1)- \pi k}$, note that $0<a_{0,n}=\sqrt \frac{2}{ \pi(4n+1)}<a_{4n,n}=\sqrt \frac{2}{ \pi}<1$. Then for all $n \in \Bbb N$ $$ \int_{[−1,1]}\vert 2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2}\vert d \mu \ge \sum_{k=0}^{4n-1} \int_{[a_{k,n},a_{k+1,n}]}\vert 2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2}\vert d \mu\\ \ge \sum_{k=0}^{4n-1} \vert \int_{[a_{k,n},a_{k+1,n}]} (2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2})d \mu \vert $$ But the integrand of $\int_{[a_{k,n},a_{k+1,n}]} (2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2})d \mu$ is Riemann integrable so we can evaluate this with Barrow's rule $$ \sum_{k=0}^{4n-1} \vert \int_{[a_{k,n},a_{k+1,n}]} (2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2})d \mu \vert= \sum_{k=0}^{4n-1}\vert a_{k+1,n}^2 \sin \frac{1}{a_{k+1,n}^2}-a_{k,n}^2 \sin \frac{1}{a_{k,n}^2}\vert=\\ \sum_{k=0}^{4n-1}\vert (-1)^{4n-k} \frac{2}{\pi}\frac{1}{4n-k}-(-1)^{4n-k+1} \frac{2}{\pi}\frac{1}{4n-k+1}\vert =\frac{2}{\pi} \sum_{k=0}^{4n-1}(\frac{1}{4n-k}+\frac{1}{4n-k+1})=\\ \frac{2}{\pi} \sum_{k=0}^{4n-1}(\frac{1}{k+1}+\frac{1}{k+2}) $$ but the armonic sum diverges, so $$ \int_{[−1,1]}\vert 2x \sin\frac{1}{x^2}−\frac{2}{x}\cos\frac{1}{x^2}\vert d \mu= +\infty $$ Similary one can prove that $$ \int_{[-1,1]}f^+= \int_{[-1,1]}f^-=+ \infty $$ However, if a function $f$ is derivable on $[a,b]$ then $f'$ is mesurable, and if $f'$ is bounded is Lebesgue integrable and $$ \int_{[a,b]}f' d \mu=f(b)-f(a) $$