As @Rammus points out this is just shorthand for taking the tensor product, rather than algebraically multiplying out as you may first assume. If we take the first term of $\phi_2$ and expand it out more explicitly with subscripts $A$ for Alice and $B$ for Bob:
$|0_A0_A\rangle(\alpha|0_B\rangle + \beta|1_B\rangle) = |0_A0_A\rangle \otimes (\alpha|0_B\rangle + \beta|1_B\rangle) = (|0_A0_A\rangle \otimes \alpha|0_B\rangle) + (|0_A0_A\rangle \otimes \beta|1_B\rangle) = \alpha|0_A0_A0_B\rangle + \beta|0_A0_A1_B\rangle$.
If we start with the original $|\phi_2\rangle$ given by e.q. 1.31
$|\phi_2\rangle = \frac{1}{2}[\alpha(|0_A\rangle + |1_A\rangle)(|0_A0_B\rangle + |1_A1_B\rangle)+ \beta(|0_A\rangle + |1_A\rangle)(|1_A0_B\rangle + |0_A1_B\rangle)]$
if we expand this out keeping track of the subscripts and remembering to keep ordering from left to right (as the tensor product is noncommutive, this means $A \otimes B \neq B \otimes A$), you can see that we will get the same terms as expanding out the rewritten form with the same ordering (grouping) of qubits for both Alice and Bob.
The second part, Alice is measuring two qubits, one which is part of her EPR pair and the other that WAS $|\phi\rangle$ however during the teleportation protocol she acts on this state and mixes this with her part of the EPR pair so we no longer have the original pure state $|\phi\rangle$. We know she can't measure $|\phi\rangle$ otherwise this would violate the no-cloning theorem as Alice and Bob would then both have a quibit in state $|\phi\rangle$ at the end of teleportation!
Daftwullie gives a great explanation here Quantum teleportation: second classical bit for removing entanglement?