Why do we assume that the wave function should satisfy the Schrödinger equation? Why do we assume that the wave function should satisfy the Schrödinger equation? If a function satisfies the Schrödinger equation, does  it mean that it is a wave function?
 A: Because wave functions satisfying the Schroedinger equation adequately describe experimental data.
I am not sure a function satisfying the Schroedinger equation is necessarily a wave function of quantum theory. For example, such a function can be used to describe Couder's experiments with bouncing droplets (https://arxiv.org/abs/1401.4356)
A: 
If a function satisfies the Schrödinger equation, does it mean that it
  is a wave function?

If, by wavefunction, you mean a position basis representation of a quantum state, then the answer is no.
For example, there are a continuum of solutions to the quantum harmonic oscillator time independent Schrödinger equation
$$\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{m\omega^2}{2}x^2\right)\psi(x)=E\psi(x)$$
but only a countably infinite subset of these solutions are normalizable and thus are wavefunction representations of states.  See, for example, this Wolfram Demonstration 
A: Because Schrödinger's equation is derived from a solution of the classical wave equation. I'm assuming this is what you mean by the wave function i.e. the vacuum solutions. That is how Schrödinger himself derived the equation. Check this paper out.
Basically, you start with the wave solutions in vacuum, substitute in the classical wave equation, and use the energy and momentum of a photon. The end result is Schrödinger's equation.
Note how the energy operator is derived from the time components, and the momentum operator from the spatial components. This fits nicely with Noether's theorem. 
