Is Earth's surface gravity termed $g$ derived from Newton's universal law of gravity? Trying to apply the basic Newton law $F = m  a$ to a falling apple and a paper clip I encountered  a problem I consider a beginner's: The fact that all objects are subject to the same acceleration termed $g$, and  falling off a balcony  arrive on the floor the very same time cannot be derived from Newtons's first three laws. Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law  that is not implied in Newton's three basic laws of motion?
If the above is correct, is $g$ ("9.x m/s2") a derivation of Newton's general law  of gravity? How exactly does this derivation look like?
Closely related:
How did Isaac Newton derive the laws of gravity?
 A: You have it backwards. The observation that objects near the Earth's surface accelerate at $g$ was one of the motivations of Newton's theory that gravity is a force proportional to mass. The known motions of the Moon and planets motivated proportionality to $1/r^2$.
In physics, we don't derive the phenomena from the theory. We find theories that capture the phenomena.
A: 
Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law that is not implied in Newton's three basic laws of motion?

Yes, that is correct. The additional law required is Newton’s law of universal gravitation. It is usually written as $$F_g=-G\frac{m_1 m_2}{r^2}\hat r$$ This additional law is not derived from the others.
A: Newton's three laws are general properties of all forces. Newton's Law of Gravitation and Coulomb's law of electrostatics are specific properties of the gravitational and Coulomb force respectively.

Is it correct to assume that the proportionality of earth's gravitation exerted on objects of different mass or weight needs some other law that is not implied in Newton's three basic laws of motion?

Yes. This is a specific property of gravity. So it's not derive-able from the three laws. The Coulomb force does not have this property

If the above is correct, is g ("9.x m/s2") a derivation of Newton's general law of gravity? How exactly does this derivation look like?

You can't derive laws, because laws are not theorems. Laws are axioms. They have to be postulated, not derived.
You have to deduce laws by observing nature. Observing that things fall at the same rate regardless of their mass helps in the deduction.
By observing that objects of different masses have the same acceleration, you deduce that force exerted by Earth on the object $F_{EO}$ is proportional to the Object's mass $m_O$.
By symmetry, the force exerted by an object on Earth, $F_{OE}$ is proportional to $m_E$.
By Newton's third law, $|F_{OE}|=|F_{EO}|=F$. So, $F$ is proportional to both $m_E$ and $m_O$, i. e. $F$ is proportional to $m_E m_O$
The dependence on $\frac{1}{r^2}$ is not observable on falling objects, because $g$ does not vary much near the surface. You have to deduce the proportionality of $F$ to $\frac{1}{r^2}$ from planetary motion observations.
