Problem with Singular Value Decomposition

You're right that the columns of $U$ and $V$ are the eigenvectors of $AA^T$ and $A^TA$ (repectively). However, that information alone does not completely determine the SVD.

Analogously: in an eigendecomposition $A = PDP^{-1}$, the columns of $P$ are eigenvectors and the entries of $D$ are eigenvalues. Nevertheless, $$ \pmatrix{0&1\\1&0}\pmatrix{1\\&2}\pmatrix{0&1\\1&0}^{-1} \neq \pmatrix{1&0\\0&1}\pmatrix{1\\&2}\pmatrix{1&0\\0&1}^{-1} $$ Once we've found a valid choice of $V$ (with eigenvectors in order corresponding to the eigenvalues of $A^TA$ in descending order), it remains to solve for a satisfactory $U$. If $A$ is invertible, then $$ A = U DV^T \implies U = AVD^{-1} $$ even if $A$ is not invertible, some of the columns of $U$ will be determined by our choice of $V$.

To put it precisely: the $i$th column of $U$ is given by $u_i = \frac 1{\sigma_i}Av_i$ whenever $\sigma_i$ is non-zero. To see that these columns are orthonormal, note that $$ u_i^Tu_j = \frac{1}{\sigma_i\sigma_j} v_i^TA^TAv_j $$

To directly answer your question, the problem is that the "typical way" you describe for getting an SVD is incorrect:

The typical way to get an SVD for a matrix $A = UDV^T$ is to compute the eigenvectors of $A^TA$ $\color{red}{and}$ $AA^T$. The eigenvectors of $A^TA$ make up the columns of $U$ and the eigenvectors of $AA^T$ make up the column of $V$.

One should either give $U$ then solve $V$, or give $V$ then solve $U$ by using the relation $A=UDV^T$.

On the other hand, you wrote that

... and thus have eigenvectors $\bigg\lbrace \left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\ ,\ \left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg\rbrace$

But one could equally have $\bigg\lbrace \left[\begin{matrix} -1 \\ 0 \end{matrix}\right]\ ,\ \left[\begin{matrix} 0 \\ 1 \end{matrix}\right]\bigg\rbrace$, or $\bigg\lbrace \left[\begin{matrix} 1 \\ 0 \end{matrix}\right]\ ,\ \left[\begin{matrix} 0 \\ -1 \end{matrix}\right]\bigg\rbrace$, or $\bigg\lbrace \left[\begin{matrix} -1 \\ 0 \end{matrix}\right]\ ,\ \left[\begin{matrix} 0 \\ -1 \end{matrix}\right]\bigg\rbrace$.

Any one of these sets consists of independent eigenvectors (with unit length) of the identity matrix. If one chooses from these sets arbitrarily to form $U$ and $V$, there is no reason to expect that $$ A=UDV^T. $$