For the matrix
$$ A=\left(\begin{array}{cc} 1 & 1\\ 1 & 1+\epsilon \end{array}\right) $$
where $\epsilon\ll 1$, the singular values are approximately $2$ and $\epsilon/2$. The first is easy to understand: we take
$$ \left(\begin{array}{cc} 1 & 1\\ 1 & 1+\epsilon \end{array}\right)\left(\begin{array}{c} 1\\ 1 \end{array}\right)\approx2\left(\begin{array}{c} 1\\ 1 \end{array}\right). $$
However, for the second, we take
$$ \left(\begin{array}{cc} 1 & 1\\ 1 & 1+\epsilon \end{array}\right)\left(\begin{array}{c} 1\\ -1 \end{array}\right)=\left(\begin{array}{c} 0\\ -\epsilon \end{array}\right). $$
The norm of the vector on the right is $\epsilon$, while that of the vector on the left is $\sqrt{2}$, so this matrix has stretched the vector by a factor of $\epsilon / \sqrt{2}$. So why is singular value $\epsilon / 2$?
Okay, I realized (with much thanks to the comments) that the key is to keep track of $\epsilon$ carefully up to second order. Let $A^{T}A=M_{0}+\epsilon M_{1}+\epsilon^{2}M_{2}$, where
$$ M_{0}=2\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right),\;M_{1}=\left(\begin{array}{cc} 0 & 1\\ 1 & 2 \end{array}\right),\;M_{2}=\left(\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right). $$
The zeroth-order eigenvalue is $\lambda_0 = 0$ and $x_{0}=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ -1 \end{array}\right)$.
At first order, we have $M_{0}x_{1}+M_{1}x_{0}=\lambda_{1}x_{0}$, leading to $x_{0}^{T}M_{1}x_{0}=\lambda_{1}$, but from a straightforward computation, this is also zero. This is the key reason why the method in the original question gives the wrong answer. Solving $M_{0}x_{1}=-M_{1}x_{0}$, we obtain $x_{1}=\frac{1}{4\sqrt{2}}\left(\begin{array}{c} 1\\ 1 \end{array}\right)$, as stated by the second comment under the original question post.
At second order, we have $M_{0}x_{2}+M_{1}x_{1}+M_{2}x_{0}=\lambda_{2}x_{0}$. Multiplying by $x^T_0$, we obtain $x_{0}^{T}M_{1}x_{1}+x_{0}^{T}M_{2}x_{0}=\lambda_{2}$. By straightforward computation, we have $\lambda_2 = 1/4$. Hence, we have $\sigma=\sqrt{\lambda_{2}\epsilon^{2}+\mathcal{O}(\epsilon^{3})}\approx\frac{\epsilon}{2}$.