Clarification of Decoding in Ring-LWE
Just so we are on the same page, I recall that the LWE problem assumes that for a secret (and usually small) vector s, it is hard to distinguish (A,b = As+e) from a random (A,b), where A is a random matrix of $\mathbb{Z}_q^{m\times n}$ and e is small. (Considerations about the required parameters and the error distribution are irrelevant for our argument, so I skip them for simplicity.)

With hypotheses for the polynomials $s,a_i,e_i$ that are similar to the ones we had for s, A, e, Ring-LWE assumes that it is hard to distinguish $(a_i,b_i = a_i\cdot s+e_i)$, which we call R-LWE samples, from random $(a_i,b_i)$.

Now, there are at least two (closely related) ways to answer your question :


The public key of Alice is $(a, b = a\cdot s+e)$, and Bob sends her an encrypted message 
$$(u,v) = (a\cdot s_b + e_b, b \cdot s_b + e_b' + m \lfloor q/2 \rceil).$$ $(a,b)$ is a R-LWE sample for the secret $s$ so it is indistinguishable from random for Eve. Similarly, if Bob sent $m=0$ to Alice, then  $(a,u) = (a, a\cdot s_b + e_b)$ and $(b,v) = (b, b \cdot s_b + e_b')$ are R-LWE samples for the secret $s_b$ so everything looks completely random to Eve and she wouldn't be able to distinguish whether Bob sent $m=0$ or random stuff to Alice.  To send $m\neq 0$, Bob adds $m \lfloor q/2 \rceil$ to $v$ and it will still look random to Eve.
I also like this more "lattice" approach. This is my interpretation and abusive notations ensue, so feel free to disagree:
$$\binom{u}{v} = \binom{a}{b}s_b + \binom{e_b}{e_b'} + \binom{0}{m \lfloor q/2 \rceil}$$
So $(u,v)$ = [point of the lattice generated by $(a,b)$] + [a small error] + [the message]. The error is chosen big enough so that Eve is unable to decode [lattice point] + [error], let alone [lattice point] + [error] + [message].
However, Alice can perform a "dot product" of $(u,v)$ with the secret vector $(-s,1)$, which is small and almost orthogonal to $(a,b)$: 
$$\binom{u}{v} \cdot \binom{-s}{1} = e\cdot s_b + (e_b'-s\cdot e_b) + m \lfloor q/2 \rceil$$
At which point Alice can easily remove the remaining errors and recover m.