Ring LWE: How can the secret be chosen from the "uniform distribution"? The key generation algorithm for Ring-LWE is as follows. Have a ring $R_p =Z_p[x]/(x^n + 1)$.
Then pick a uniformly random $a$ from $R_p$.
Pick $s$ from an appropriate distribution.
Pick $e$ from the error distribution (typically a constant-size Gaussian).
Compute $b = as + e$. Reveal $(a, b)$ as the public key and $s$ is the secret key.

I've been reading The Homomorphic Encryption Standard. It says that $s$ can be picked as a uniform random element from $R_p$. How is this possible? For instance if the (admittedly insecure) choice $n=1$ was picked, then for any values of $a$, $b$ and $e$ there is some value $s$ which satisfies the equation.
More seriously, the LPR scheme, for instance, assumes that $(re - se_1 + e_2)$ is small in order for decryption to be correct. How can it be small if $s$ is chosen uniformly at random from $R_p$?
You're right that essentially all schemes require that the secrets are small in a concrete sense.
While I'm not terribly familiar with the standard, I believe the tables in question were generated by the LWE Estimator of Albrecht et al. This page of its documentation states various secret distributions they support. I expect "uniform" corresponds to "bounded uniform", but I'm unsure of which particular bound was used to generate that table.