Is my simplistic understanding of a Qubit, Superposition and Entanglement correct? I've gone through a number of Lenny Susskind's lectures on entanglement (both just for the joy of it and to better understand quantum mechanics) where he delves into the idea of a qubit.
For the sake of this example, let's go with the magnetic moment of an electron which can be 0 or a 1 at measurement. I'll go with 2 qubits alongside the classical model as a comparison.
Classical Bit Model
A two bit system (00, 01, 10, 00) has for possibilities and can be described using 2 bits of information.
Qubit Model
The Qubit model consists of the following:
a|00>
b|01>
c|10>
d|11>

a|00> + b|01> + c|10> + d|11> = 1

Where a|00>, b|01>, c|10>, and d|11> squared are complex amplitudes representing the probabilities of each state at measurement respectively. In this case, there are 4 bits of information required to describe this system.
Subsequently, a 3 bit system would require 8 bits of information (2^3) to describe.
Entanglement refers to the fact that both bits are described by one wave function when calculating the probabilities of each outcome. They no longer vary independently.
Superposition refers to the idea that the system is a linear combination of each state (akin to northeast in comparison to north and east separately) until the system is measured; after which it collapses into a single defined and possible state.

The classical model requires n bits to describe a system of n bits, whereas a quantum mechanical model requires 2^n bits to describe a system of n bits.
Thus, a quantum computer requires an exponential amount of information to describe its state in comparison to an equivalent classical computer of the same bit number.
 A: There's some problems with your suppositions:
$$a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle= \begin{pmatrix} a \\ 
b \\
c \\
d\end{pmatrix}$$
Which are just two different notations -- the complex linear combination is not simply 1. There's the Measurement Postulate, which would have that, upon measurement, the probability of measuring the pair in any particular state is $1.0$, given that a measurement had taken place -- but this is not simply the natural state of things.
$a,b,c,d$ are probability amplitudes, $a, b \in \mathbb{C}$ and $|00\rangle, |01\rangle,$ etc. form a basis representing the combination of two qubits in a pure state.
I'm not sure where you are gleaning these information quantities from. There's one analog I know of called superdense coding - a protocol with which one qubit in an entangled pair can be used to transmit two bits of information to a recipient. I know of no other metric for quantifying the amount of information contained in any quantum state.
You are correct, in the established notation. Entanglement is when you have a 2-qubit state that cannot be expressed in the computational basis, because a pair of arbitrary states taken together has, $$(a|0\rangle +b|1\rangle) \otimes (c|0\rangle + d|1\rangle)=\begin{pmatrix} ac \\ ad \\ bc \\bd \end{pmatrix}$$
and an operation like swapping $bc$ with $bd$ (which would correspond to $\text{CNOT}$, a controlled-NOT gate) can be constructed, but then the new state can no longer be expressed as a tensor product, like how we constructed it.
Your statement regarding superposition is correct.
There's no correct analog though with respect to digital information contents, though:
$$|\psi\rangle =\begin{pmatrix} a \\ b\end{pmatrix}$$ exists in a continuous Hilbert space, when discussing universal quantum computing.
A: The answer by meltyness is nice, and I would upvote it if I had rep here. Here are some more general and less mathematical ideas that may help you to look at this from a different angle.
Classical and quantum information are fundamentally different things, and they're not freely interconvertible. Quantum information is a higher grade of information. It's sort of like in thermodynamics, where mechanical work is a higher grade of energy than heat. You can convert work into heat, but you can never completely convert heat into work. A more formal way to state this is that you can't in general transmit quantum information over a classical channel.
A good way to see this is to think about the fact that you can take a qbit and smoothly transform it from one basis state (say spin up) to another (say spin down), and you can do this through a transformation that involves only pure states (i.e., we're not talking about ensembles of electrons but just one electron). How would you represent this continuous change using any finite number of classical bits? A classical memory can only be in a finite number of states.
We can also get at this through the no-cloning theorem. It's not possible to make a separable copy of an unknown, pure quantum state. Doing so would violate unitarity, since it would erase the information of the system being overwritten with the copy. Well, if you could transform quantum information losslessly into classical information and then back, then you could just make the classical version, copy it as many times as you liked, and then transform the copies back to quantum -- which would violate no-cloning.
It's really not that surprising that we can't read out a quantum state as classical information. If we could, then we could get absolute phase information about a system. For instance, I could read out the state of a hydrogen atom as quantum information, convert it to classical, and write down the whole thing, including phase, on a piece of paper. Then I could wait some amount of time while the phase spun around as required by the Schrodinger equation, $E=\hbar\omega$, and record the new phase. But that doesn't make sense. The energy E is only defined up to an additive constant. The only way to get information about a phase like this is to interfere my hydrogen atom with another hydrogen atom, which gives only probabilistic results about their relative phases.
When you read out results from a quantum computer, you're doing a measurement. Due to decoherence, all you get from a measurement is an eigenvalue of the observable you measured. If you measure a qbit to be spin-up, then that's all you get. You can't tell what was the amplitude of the spin-down part before the measurement or the relative phases of the two parts of the superposition.
