Why do we choose the smallest singular value? What do the other singular values represent?

Also, is it $ x' \equiv H x $ (i.e. only up to a scale)?

Let's say we have the SVD of a matrix $A = U \Sigma V^T$, where $u_i$ represents the $i$-th column of $U$ (left singular vector), $v_i$ represents the $i$-th column of $V$ (right singular vector), and $\sigma_i$ is the $i$-th singular value (a diagonal element of $\Sigma$). Then, one can show that $u_i^T A v_i = \sigma_i$.

Suppose the $i$-th singular value is $0$, where $u_i^T A v_i = 0$. Since both $u_i \neq 0$ and $v_i \neq 0$, then it follows that both $A v_i = 0$ and $A^T u_i = 0$ must be true. And therefore, $x = v_i$ solves the homogeneous equation $Ax = 0$ used to recover the homography $H$.

And yes to your second question: $x' = s H x$ for some $s \neq 0$.

thank you! so if the singular value is not 0 (perhaps due to noise or poor correspondence), then the smallest singular value (i.e. that closest to 0) would be the best estimate for $h$?

Yup! Exactly. The singular vector would minimize the value of $|A v_i|$ subject to $|v_i| = 1$.