I read the following in a paper and I have a hard time understanding it.
... unit norm eigenvector are at most defined upto a single sign flip and we have to choose the signs synchonously.
What does "unit norm eigenvectors are defined unto a single sign flip" mean? Does it mean that "if $v_i$ is an eigenvector then $-v_i$ is also an eigenvector?
You're right: "if $v_i$ is an eigenvector then $−v_i$ is also an eigenvector".
Indeed, eigenvectors of a matrix $A$ are specified by $$ A v_i = \lambda_i v_i $$ with eigenvalues $\lambda_i$. So with any constant $c$, $c v_i$ also satisfies the equation.
It is usual to choose $c = d/|v_i|$ with $|d| = 1$ so you have unit norm eigenvectors. However, this still leaves you the choice of $d$ with $|d| = 1$. Not only can you choose $d = \pm 1$, but also complex values with norm 1, in case the solutions are not real.