What does one mean by saying that classical bits perform operations at the scale of $2n$ and quantum computers perform operations at the scale of $2^n$? In both cases, $n$ = Number of bits/qubits.
I'm not sure it really is true to make such a claim, even though it is one that is often seen. Even so, this statement is common because it does point towards a difference between classical computers and quantum ones.
Classical computation is essentially a process that takes a single input bit string and keeps transforming it until you get a single output bit string. You can think of the whole process as only ever having one bit string in the computer at once.
The same is true for quantum computers, except that you need to replace 'bit string' with 'state of many qubits'. So how do bit strings compare with multi qubit states? To find out, we can look at how classical computers can simulate quantum ones.
One way to represent states of $n$ qubits in a classical computer is to think of them as superpositions of all possible $n$-bit strings. Then you can have a big array, which stores the corresponding amplitude for every $n$-bit string. Since there are $2^n$ $n$-bit strings, this will take an exponentially large amount of memory. This is often not the most efficient method of representing quantum states, but there are cases when it is no worse than any other.
So if we need to simulate any possible process for $n$ qubits with bits, we know that it will incur this kind of overhead. If we use this to draw a comparison between qubits and bits, we could say that an exponentially large number of qubits are required to match the power of $n$ bits. But the same would not be true for all possible computational tasks.
The reason is because of the superposition. It allows you to perform operations and won the speed up. For example, if you have a just one qubit you will have the following because of the superposition:
$$\alpha_0\lvert0\rangle + \alpha_1\lvert1\rangle$$
You can see that you have already $2$ basis for one qubit. If you have two qubits you will have $4$ basis for your system and go on. I will put the $2$ qubit but as a combination of tensor product of the previous basis and we will have:
$$\lvert0\rangle ⊗ \lvert0\rangle, \lvert0\rangle ⊗ \lvert1\rangle, \lvert1\rangle ⊗ \lvert0\rangle, \lvert1\rangle ⊗ \lvert1\rangle$$
If you apply the tensor product of it, you will end up with a $4$ basis system for $2$ qubits ($2^2$).
If you are looking for a introduction material I think that the Lecture Notes from Ronald de Wolf is a good start. It is possible to get directly from his website and it is for free. He gave a better explanation about this on section 1.3.
I hope that I have helped you.
By saying that a quantum computer using $n$ qubits does (up to) $2^n$ computations in parallel, one tries to explain quantum parallelism: If you represent the state of the $n$ qubits using probability amplitudes for each state in the computational basis, there are $2^n$ such probability amplitudes that a classical computer would have to update per quantum gate it is to simulate, whilst a quantum computer does this automatically (but with potentially less benefit since not all these numbers can be independently measured).