States and observables in quantum mechanics I'm beginning learn quantum mechanics. As I understand, state is a map $\phi$ from $L^2(\mathbb R)$ such that $|\phi|^2$ describes probability density of a particle's position. By integrating $|\phi|^2$ we can get the probability of a particle's position in a subset of $\mathbb R$.
My question is. Why we need $\phi$? Why just not speaking about probabilities itself, i.e. probability measures $\mathsf \Phi$ on $\mathbb R$ such that $\Phi(A)$ is the probability of particle's position in $A\subset \mathbb R$.
Moreover, why we speaking about observables as operators. Why not just speaking about random variables on $\mathbb R$. (I know that observable isn't a arbitrary operator; but may be the restrictions can be expressed in terms of random variables?)
Many expressions in quantum mechanics via integrals with observables and states, as I understand, just express expectation, variation and standard deviation of such random variables w.r.t. probability that define the states. For example, Heisenberg's inequality just inequality for product of standard deviations of position and momentum.
 A: Why we need ϕ$\phi$? Why just not speaking about probabilities itself,
Because the wave function is a complex function and contains more information than the (real) probability density. And we NEED this information, i.e. there are observables that use it (see below). 
Many expressions in quantum mechanics via integrals with observables and states, as I understand, just express expectation, variation and standard deviation of such random variables w.r.t. probability that define the states.
Many, but not all. The expectation value of momentum is, for instance, the average of $\phi^* \phi'$, which can't be calculated from the probability density. 
A: 
My question is. Why we need $\phi$? Why just not speaking about probabilities itself, i.e. probability measures $\mathsf \Phi$ on $\mathbb R$ such that $\Phi(A)$ is the probability of particle's position in $A\subset \mathbb R$.

One of the principles of quantum mechanics is that position and momentum are not independent but do not determine one another. This can be made precise in various ways, e.g., $[\hat{x},\hat{p}] = i\hbar$ means the position-space wavefunction is related to the momentum-space wavefunction through a Fourier transform, which instantiates the de Broglie relations. The phase of a particle's position-space wavefunction is not irrelevant--as a consequence of the Schrödinger equation, it gives information about momentum, related to the gradient of the phase. 
In Quantum Mechanics and Path Integrals, Feynman and Hibbs expressed the view that quantum mechanics represents a replacement of probability theory itself. That's probably a good way of thinking about it: you're still dealing the "probabilities [themselves]", but they behave in a blatantly non-classical way--for the wavefunction representation, they add like complex numbers.
You don't actually 'need' a wavefunction, but it encapsulates the probabilities of all possible measurements in straightforwardly (assuming a pure state). As an alternative, Wigner quasiprobability distributions work in an analogue of a classical phase space, but they can assign negative probabilities (for incompatible measurements, fortunately). Thus, you could work with distributions that add in a classical manner if you wanted to, at the price of departing from standard probability theory in yet another way.
But whatever you do, you're not going to fit quantum mechanics in a classical probability theory. It's clear from the double-slit experiment that something very wonky is going on, and Bell's inequalities can prove this in general. Wavefunctions are not your only choice, but classical probabilities don't work.

Moreover, why we speaking about observables as operators. Why not just speaking about random variables on $\mathbb R$. (I know that observable isn't a arbitrary operator; but may be the restrictions can be expressed in terms of random variables?)

Again, you could talk about random variables, as Wigner distributions do this, but not for individual observables (that wouldn't describe the state fully), but being quasiprobabilities, they don't fit into standard measure theory. As for why observables are operators, it's because they form an algebra that can be represented via operators on a Hilbert space. Perhaps surprisingly, there is nothing quantum-mechanical in that statement.
Let's back up a bit. The classical phase space describes the state as a combination of positions an momenta, $(q;p)$. Observables are continuous real-valued functions over the phase space, e.g., position $q$, momentum $p$, kinetic energy $T(q;p) = p^2/2m$, etc., with the time evolution of any observable $f$ being given by the Poisson bracket with the Hamiltonian:
$$\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial t} + \{f,\mathcal{H}\}\text{.}$$
These observables can be added, multiplied, and so forth, and in general form a Poisson algebra with the Poisson bracket as the Lie bracket. Even in classical mechanics, observables form an algebra, a commutative one.
By the Gel'fand-Naĭmark theorem, a somewhat general class of algebras ($C^\star$) can be represented as the algebra of bounded linear operators on some complex Hilbert space. (Although since in most situations the interesting observables in QM aren't bounded, how this is generalized to include unbounded operators is somewhat hairy; some of textbooks, e.g., Griffith's, just put a blurb in a footnote about rigged Hilbert spaces, a decision I'm going to emulate here.)

Many expressions in quantum mechanics via integrals with observables and states, as I understand, just express expectation, variation and standard deviation of such random variables w.r.t. probability that define the states

None of these things are inherently quantum-mechanical.
As implied by the previous section, even classical mechanics can be formulated as having the physical state as a vector on a complex Hilbert space, observables as (some) linear operators with measurements outcomes being the eigenvalues. There state can be represented as a complex wavefunction $\Psi(x,p;t)$ with measurement collapsing it to the eigenspace corresponding to the result, and it evolves in by an analogue Schrödinger equation (the time-evolution given by the Liouvillian operator sintead). This has been done in detail by Koopmnan and von Neumann.

For example, Heisenberg's inequality just inequality for product of standard deviations of position and momentum.

Now that's very quantum-mechanical. The Heisenberg uncertainty principle is a direct consequence of non-commutativity of observables:
$$\sigma_A\sigma_B \geq \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle\right|\text{,}$$
where $[\cdot,\cdot]$ is the commutator.
A: The reason for that is that there is no differential equation which governs the probability density, only one which governs the 'wave function'. Remember that the wave function is a complex valued function so there is no one to one correspondence between the wave function and the probability density. Given a particular probability density, the value of the wave function can lie anywhere on the corresponding circle in the complex plane. If you consider for eg., the Schrodinger equation for the free particle in one dimension:
$$ \frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2\psi}{{\partial x}^2} $$
and write $\psi$ as:
$$ \psi = Re^{i\Theta}$$, you'll find that the terms containing the derivatives of $R$ and $\Theta$ are coupled and cannot be separated into two separate differential equations.
Can't answer the other part of your question as don't know what you mean by 'random variable'.
