The choice of eigenvectors in the SVD To find the Singular Value Decomposition $A=U\Sigma V^\dagger$ of
$$
A = \begin{pmatrix}
0 & -1 \\
1 & 1 \\
-1 & 0
\end{pmatrix}
$$
I first identified the eigenvectors of $AA^{\dagger}$ (which will be the columns of $U$). The eigenvectors corresponding to the eigenvalues $\lambda_1 = 3,\ \lambda_2 = 1, \ \lambda_3 = 0$ are chosen, so that you get an orthonormal set, to be
$$
u_1=\frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\-2 \\ 1\end{pmatrix},\
u_2=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\0 \\ -1\end{pmatrix},\
u_3=\frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\1 \\ 1\end{pmatrix},
$$
which makes
$$
U =\frac{1}{\sqrt{6}}\begin{pmatrix} 1 & \sqrt{3} & \sqrt{2} \\
                  -2 & 0 & \sqrt{2}\\
                   1 & -\sqrt{3} & \sqrt{2} 
\end{pmatrix}
$$
and
$$
\Sigma = 
\begin{pmatrix}
\sqrt{3} & 0 \\
0 & 1 \\
0 & 0
\end{pmatrix} .
$$
Similarly,
$$
V = \frac{1}{\sqrt{2}}\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}.
$$
It seems to me that, up to this point, I have carefully followed the procedure of SVD method. However,
$$
U\Sigma V^\dagger =
\begin{pmatrix}
1 & 0 \\
-1 & -1 \\
0 & 1
\end{pmatrix} 
$$
which is not $A$; the sign is wrong and the columns are switched. After further calculation I found that if I flip the sign of $u_1$, I get $\ U\Sigma V^\dagger = A$. But what have I done wrong? Is there a way to tell if a given choice of eigenvectors is the “right” set of eigenvectors (“right” in the sense that I get $\ U\Sigma V^\dagger = A$)?
 A: I assume by "similarly", you mean to diagonalize the matix $A^{\dagger}A$. Because there is some flexibility in picking the unit eigenvectors up to a sign (as there are only two elements in $\{x\in\mathbb R \mid |x|=1\}$ and it's much worse if we work with $\mathbb C$ instead of $\mathbb R$), you won't be able to coordinate $U$ and $V$ well in this way, and also this involves some unnecessary computations.
After you get $U$, to get $V^{\dagger}$ is straightforward: $A=U\Sigma V^{\dagger}$, hence $U^{\dagger}A = \Sigma V^{\dagger}$. The LHS can be computed directly as $U$ hence $U^{\dagger}$ has been calculated. To get $V^{\dagger}$, we compare the columns on both sides: If $\Sigma$ has a nonzero value on the $i$-th column, then the $i$-th column of $V^{\dagger}$ is just the corresponding column of $U^{\dagger}A$ divided by $\Sigma_{ii}$; after all these columns are done, the rest of $V^{\dagger}$ will be forming an orthonormal complement basis of the scalar multiples of $U^{\dagger}A$, which can be obtained through some Gram-Schmidt process.
In your particular case, there is no need to do the Gram-Schmidt, as $V^{\dagger}$ has exactly two columns and $\Sigma$ happens to have two nonzero (diagonal) elements.
