As I suggest in the comments, I don't think that it is going to help you to understand quantum computation in terms of parallelism. To illustrate why, I will describe a simple two-qubit computation, in which — if you were absolutely adamant — you could claim there is computation happening in parallel; but which I would suggest does not in any meaningful sense.
$\def\ket#1{\lvert#1\rangle}$
Consider the following circuit, acting on some standard basis state $\ket{x}\ket{y}$ provided as input:
To smooth the analysis slightly for those who haven't seen a lot of quantum computation, let's consider how we could most easily represent the effects of these operations on some states.
The middle gate, which is a $\mathrm{CNOT}$ gate ("controlled-not"), performs the following transformation of standard basis states:
$$ \mathrm{CNOT} \ket{x}\ket{y} = \ket{x}\ket{y \!\oplus\! x}$$
and performs the same transformation, independently, on each term of a superposition of standard basis states. This aspect of performing things independently on individual terms is what is sometimes described as the 'parallel' behaviour of quantum computation.
The gates surrounding the $\mathrm{CNOT}$ gate in the circuit are Hadamard gates,
$ H = \tfrac{1}{\sqrt 2}\bigl[\begin{smallmatrix} 1 & \phantom- 1 \\ 1 & -1 \end{smallmatrix}\bigr], $
which we may describe as performing the following transformation on standard basis states:
$$ \ket{x} \;\xrightarrow{\;H\;}\; \tfrac{1}{\sqrt 2}\Bigl( \ket{0} + (-1)^x \ket{1} \Bigr) .$$
On pairs of standard basis states, we may represent the effect of Hadamards on both qubits by
$$ \ket{x}\ket{y} \;\xrightarrow{\;H \otimes H\;}\; \tfrac{1}{2}\Bigl( \ket{00} + (-1)^y \ket{01} + (-1)^x \ket{10} + (-1)^{x \oplus y} \ket{11}\Bigr); $$
and because the Hadamard gate is self-inverse, we also have the reverse transformation of states,
$$ \tfrac{1}{2}\Bigl( \ket{00} + (-1)^y \ket{01} + (-1)^x \ket{10} + (-1)^{x \oplus y} \ket{11}\Bigr) \;\xrightarrow{\;H \otimes H\;}\; \ket{x}\ket{y}. $$
So: given these observations, let's take a look at what happens when we perform the circuit illustrated above on an input state $\ket{x}\ket{y}$: reading off transformations of the state, time-step by time-step, we have
$$
\begin{align}
&
\ket{x}\ket{y}
\\[1ex]&\xrightarrow{\;H \otimes H \;}
\tfrac{1}{2}\Bigl( \ket{00} + (-1)^{y} \ket{01} + (-1)^x \ket{10} + (-1)^{x \oplus y} \ket{11}\Bigr)
\\[1ex]&\xrightarrow{\;\mathrm{CNOT} \;}
\tfrac{1}{2}\Bigl( \ket{00} + (-1)^{y} \ket{01} + (-1)^x \ket{11} + (-1)^{x \oplus y} \ket{10}\Bigr)
\\&\qquad\qquad=
\tfrac{1}{2}\Bigl( \ket{00} + (-1)^{y} \ket{01} + (-1)^{x\oplus y} \ket{10} + (-1)^{x} \ket{11}\Bigr)
\\&\qquad\qquad=
\tfrac{1}{2}\Bigl( \ket{00} + (-1)^{y} \ket{01} + (-1)^{(x\oplus y)} \ket{10} + (-1)^{(x\oplus y) \oplus y} \ket{11}\Bigr)
\\[1ex]&\xrightarrow{\;H \otimes H \;}
\ket{x \!\oplus\! y}\ket{y}.
\end{align}$$
In the middle of the computation, we have a superposition of standard basis states, and $\mathrm{CNOT}$ can be said to be operating independently on each of them. But if we were to describe this as "parallel computation", you should ask yourself:
- What is the data that the computation is acting on?
- What is the output of these 'parallel' processes?
It seems to me that the standard basis terms that the $\mathrm{CNOT}$ is 'acting' on, aren't data at all: they do not individually correspond to any information about the input state. Furthermore, information about individual terms in the computation aren't represented in the output. It is customary — and in some contexts, essentially correct — to describe the terms in the superposition as representing random bits (or bit-strings), and you could say that it is on this random information that the process acts in parallel; but this 'randomness' does not in any way affect the output.
What this example demonstrates is that quantum computation allows information to be stored in different ways (using different orthogonal bases) of a single-qubit or multi-qubit system, and that the way in which operations act on these different bases is determined by — but may have qualitatively different features from — the way it acts on the standard basis. It's possible to describe parts of computation as acting 'in parallel over different possibilities', but such descriptions shouldn't be taken too seriously, because the 'parallelism' is one that one cannot directly exploit, any more than you can exploit the similar 'parallelism' that exists in randomised algorithms.
