Let us work through your example, i.e., $2^6$-dimensional normalized complex vector $|\psi\rangle$
\begin{equation} |\psi\rangle=\sum_{\sigma_1,\sigma_2,\ldots,\sigma_6}\psi_{\sigma_1,\sigma_2,\ldots,\sigma_6}|\sigma_1,\sigma_2,\ldots,\sigma_6\rangle
\end{equation}
The matrix $\psi_{\sigma_1,\sigma_2,\ldots,\sigma_6}$ has shape $(1,2^6)=(1,64)$. This is equivalent to thinking of $\psi$ as being a $(2,2,2,2,2,2)$ tensor. Visually we can think of this tensor as a block with 6 legs, where each of the legs can take in two (or d- in the general case) values - corresponding to the values of $\sigma_i$ - here 0 or 1. This is the situation in the top row of the attached Figure 5 from the review you sent.
The goal now is to work through the block from left to right and decompose our state into a product of smaller tensors (big dots in the bottom row of the figure) - or more formally:
\begin{equation}
\psi_{\sigma_1,\sigma_2,\ldots,\sigma_6} = \sum_{a_1,a_2,\ldots,a_5} A^{[\sigma_1]}_{a_1}A^{[\sigma_2]}_{a_1,a_2}\ldots A^{[\sigma_5]}_{a_4,a_5} A^{[\sigma_6]}_{a_5}
\end{equation}
Note that the $\sigma_i$ indices correspond to the edges extending up (these are the physical indices corresponding to the values of $\sigma$ at each site), while the $a_i$ indices correspond to the horizontal bonds (these are the bond indices - which aren't physical and depend on the way we construct the MPS). Additionally, note that the first and last sites have only one bond index ($a_1$ and $a_5$, respectively).
The meaning of "decomposing $U$ into $d$-row vectors" should now become clearer: For each of the $d$ choices of the value of the physical index $\sigma_i$, we have a matrix of shape $(\dim(a_{i-1}),\dim(a_i))$. In the edge cases (first and last step), matrices $A^{[\sigma_1]}$ and $A^{[\sigma_6]}$ are just vectors (for fixed values of $\sigma_1,\sigma_6$).
The $U$ matrices will have the shapes (2,2), (4,4), (8,8), (16,4), (8,2), (4,1), and they correspond to the whole $i$-th vertex in the figure. Note here that because after the third step, matrix M (on which we do SVD) has dimensions $m\times n$ with $n<m$, U no longer becomes a square matrix; see Eq. 18 in the review.
To get the form of matrices $A$, we now want to decouple the physical indices from the bond indices of the $U$ arrays. That is, for the collection of the $2$ (or $d$ in general) values of the physical index $\sigma_i$, we have matrices $A^{[\sigma_i]}$ and their shapes are (1,2), (2,4), (4,8), (8,4), (4,2), (2,1).
The matrix elements of A's are then related to U's exactly as described in the review, namely
\begin{equation}
A^{[\sigma_1]}_{a_1} = U_{\sigma_1,a_1} = \langle \sigma_1|U|a_1\rangle
\end{equation}
for the edge cases (that is just like taking the row vectors from the 2x2 matrix) or
\begin{equation}
A^{[\sigma_2]}_{a_1,a_2} = U_{(a_1\sigma_2),a_2} = (\langle a_1| \otimes \langle \sigma_2|)U |a_2\rangle
\end{equation}
Note that the $(a_1\sigma_2)$ works in the sense of a tensor product (added braket notation for clarity).
To summarise I will show the procedure step-by-step, as to how the shapes of our state representation change over time (short-arrow denotes reshaping from $U\to A$, Long-arrow denotes reshaping of the state):
- Starting point: $(1,2^6=64)\longrightarrow(2,32)$
- $U=(2,2)\to A^{[\sigma_1]}=(1,[2],2)$, $SV^\dagger = (2,32)\longrightarrow(4,16)$
- $U=(4,4)\to A^{[\sigma_2]}=(2,[2],4)$, $SV^\dagger = (4,16)\longrightarrow(8,8)$
- $U=(8,8)\to A^{[\sigma_3]}=(4,[2],8)$, $SV^\dagger = (8,8)\longrightarrow(16,4)$
- $U=(16,4)\to A^{[\sigma_4]}=(8,[2],4)$, $SV^\dagger = (4,4)\longrightarrow(8,2)$
- $U=(8,2)\to A^{[\sigma_5]}=(4,[2],2)$, $SV^\dagger = (2,2)\longrightarrow(4,1)$
- $U=(4,1)\to A^{[\sigma_6]}=(2,[2],1)$
Dimension of the physical index $\sigma$ in square brackets.
