In R-LWE, is there any advantage to generate secret from normal distribution instead of uniform distribution
Neither the secret nor the error in Ring-LWE (or LWE) encryption needs to be generated from the normal distribution.  Only to get the tightest security proofs does one care about the "shape" of the distribution.  For practical cryptanalysis, this does not matter -- only the size of the secret (and noise) does.

Having said this, there is an advantage to the normal distribution.  During decryption, terms that look like $\langle s,e\rangle$ (inner product of the secret vector $s$ and error vector $e$) appears in the decryption.  What we want for security is that $s$ and $e$ have large norms.  What we want to avoid decryption errors is that $\langle s,e\rangle$ is small.  And if one wants to have vectors $s$ and $e$ of a particular $l_2$ norm and have their inner-product be as small as possible, then these vectors should be chosen from a normal distribution.  

The binomial distribution is a very good approximation to the normal and is much easier to generate.  This is why it's a good choice for practical applications.   

Edit:
My original answer was incomplete.  Actually, the uniform distribution is just as good as the binomial one if one only cares about the above two properties.  The main advantage of the binomial distribution is that it has more entropy.  This is important to prevent brute force and man-in-the-middle attacks.