Is the tensor product of two states commutative?

The idea of a tensor product is to link two Hilbert spaces together in a nice mathematical fashion so that we can work with the combined system. Normally, these two Hilbert spaces each consist of at least one qubit, and sometimes more.

Let's say we have a qubit, which we label $a$, and a qubit which we label $b$. These qubits 'live' in the Hilbert spaces of $mathcal{H}_{a}$ and $mathcal{H}_{b}$, respectively; we might call their respective states $|psi_{a}rangle$ and $|psi_{b}rangle$. The idea of the tensor product is that we can write the state of the two system together as: $$|psi_{ab}rangle = |psi_{a}rangle otimes |psi_{b}rangle.$$ We have 'linked' the Hilbert spaces $mathcal{H}_{a}$ and $mathcal{H}_{b}$ together into one big composite Hilbert space $mathcal{H}_{ab}$:

$$ mathcal{H_{ab}} = mathcal{H}_{a} otimes mathcal{H}_{b}. $$

Of course, there is no reason that qubit $a$ should come before qubit $b$. We thus also could link their Hilbert spaces together in reversed order:

$$ mathcal{H_{ba}} = mathcal{H}_{b} otimes mathcal{H}_{a}. $$ We need to respect our new ordering, and therefore the state of the two systems together is now:

$$ |psi_{ba}rangle = |psi_{b}rangle otimes |psi_{a}rangle. $$

Mathematically speaking, this is a different vector than $|psi_{ab}rangle$. This is exactly because we have rearranged the order of qubits in how we linked them together.

Explicit example

Let's say that we have a qubit $a$ in the Hilbert space $mathcal{H}_{a}$ with the state $$|psi_{a}rangle = alpha |0_{a}rangle + beta|1_{a}rangle,$$ and a qubit $b$ in the Hilbert space $mathcal{H}_{b}$ with the state $$|psi_{b}rangle = gamma |0_{b}rangle + delta|1_{b}rangle.$$

We can link these two qubits together with $a$ first: $$ |psi_{ab}rangle = alphagamma |0_{a}0_{b}rangle + alphadelta |0_{a}1_{b}rangle + betagamma |1_{a}0_{b}rangle + betadelta|1_{a}1_{b}rangle, $$ where I now have specifically labeled the basis states for qubit $a$ and $b$.

Or with $b$ first: $$ |psi_{ba}rangle = alphagamma |0_{b}0_{a}rangle + betagamma|0_{b}1_{a}rangle + alphadelta |1_{b}0_{a}rangle + betadelta|1_{b}1_{a}rangle. $$

These state are not the same. We see that the coefficients for $|01rangle$ and $|10rangle$ have been interchanged, but why this happened becomes very obvious if we look at the labels $a$ and $b$ of the basis states. All we have done is writing $a$ or $b$ first.

As an added argument, you could have the SWAP operation act on either of these states, and arrive at the other one. Note that, if we are very scrupulous, strictly speaking, by applying the SWAP gate we have not (re)reversed the order, but we have just 'given' the state of qubit $a$ to qubit $b$ and vice versa. If you may, it is kind of like a 'double fault', that cancels itself out.

So in general a tensor product does not commute, but rearranging the terms is just reordering the systems that you link. We just stick with one particular ordering, and it is always evident which one this is.

Annoyingly, the answer is "it depends on what you mean by $|psirangle otimes |phirangle$."

A tensor product of vectors from a collection of Hilbert spaces over the same field is simply a choice of one vector from each Hilbert space, with some equivalence relations modded out. (A tensor product of Hilbert spaces is more complicated.) Let $mathcal{H}_A$ and $mathcal{H}_B$ be the Hilbert spaces for two distinct physical systems $A$ and $B$ (e.g. qubits), and let $|psirangle in mathcal{H}_A$ and $|phirangle in mathcal{H}_B$. Then $|psirangle_A otimes |phirangle_B$ and $|phirangle_B otimes |psirangle_A$ are indeed identical vectors; clearly the choice itself doesn't depend on the order in which you write down its pieces.

But if there is a natural isomorphism between $mathcal{H}_A$ and $mathcal{H}_B$ (which is the case by definition if the two systems are "physically equivalent" at the level of our quantum description), with $(|psirangle in mathcal{H}_A) equiv (|psi'rangle in mathcal{H}_B)$ and $(|phirangle in mathcal{H}_B) equiv (|phi'rangle in mathcal{H}_A)$, then $|psirangle_A otimes |phirangle_B$ is definitely not necessarily equal to $|phi'rangle_A otimes |psi'rangle_B$. The tensor product doesn't even "know about" our isomorphism between the factor spaces, so there's no way that this relationship could hold. These two vectors correspond to the two (equivalent but distinct - a subtle concept!) systems being assigned the same two physical states, but swapped.

Without Hilbert space subscripts, the notations $|psirangle otimes |phirangle$ and $|phirangle otimes |psirangle$ are somewhat ambiguous, so the notation is underspecified for determining whether or not they're identical. Your author is interpreting the Hilbert space label as being "pinned" to the particular vector, so it swaps along with the symbols $|psirangle$ and $|phirangle$. In this case, indeed $|psirangle otimes |phirangle equiv|phirangle otimes |psirangle$.

However, physicists often use the following notational convention:

1. If there exists a natural physical isomorphism between Hilbert spaces $mathcal{H}_A equiv mathcal{H}_B$, then use the same notation for isomorphic vectors between the Hilbert spaces, with possibly only a subscript distinguishing them. That is, if $(|psirangle in mathcal{H}_A) equiv (|psi'rangle in mathcal{H}_B)$, then simply use the notations $|psirangle_A$ and $|psirangle_B$ to describe them.
2. Fix the order of factor Hilbert spaces in a tensor product once at the beginning of the discussion, and then use that order consistently throughout the entire discussion. This allows you to drop Hilbert space subscript labels and save writing.

Under this second convention (which is more common with physicists), we have $|psirangle otimes |phirangle notequiv|phirangle otimes |psirangle$, as explained in the third paragraph.

What the author is trying to say is that it does not matter in which order you write down the two subsystems, it is still the same state. It becomes much clearer if you add subscripts for the subsystems $1$ and $2$. Let's say $H_1$ and $H_2$ are finite dimensional with bases $|u_irangle_1$ and $|v_jrangle_2$. The Hilbertspace $H_1otimes H_2$ has a basis $|w_{ij}rangle = |u_irangle_1 otimes |v_jrangle_2$, but the order does not matter: $|w_{ij}rangle=|u_irangle_1 otimes |v_jrangle_2 = |v_jrangle_2 otimes |u_irangle_1$. The same applies to the states $|phirangle$ and $|chirangle$: $$ begin{aligned} |phirangle_1 otimes |chirangle_2 =& sum_{ij} phi_i chi_j |v_irangle_1otimes |u_jrangle_2 =& sum_{ij} phi_i chi_j |w_{ij}rangle =& sum_{ij} phi_i chi_j |u_jrangle_2otimes |v_irangle_1 =& |chirangle_2 otimes |phirangle_1 end{aligned} $$ Your confusion probably comes from the fact that the coefficients in a vector representation change when you change the basis. This is because in general the tensor product is not commutative. For example: $$left( begin{array}{c} 1 0 end{array}right) otimes left( begin{array}{c} 0 1 end{array}right) = left( begin{array}{c} 01 0 0 end{array}right)$$ but $$left( begin{array}{c} 0 1 end{array}right) otimes left( begin{array}{c} 1 0 end{array}right) = left( begin{array}{c} 0 1 0 end{array}right).$$