For a $general$ triple cover $f \colon X \to Y$ the situation is as follows. 

Let  $R \subset Y$ be the ramification divisor and $B \subset Y$ the branch divisor, that is $B = f(R)$. Then $R$, $B$ are both reduced and irreducible, and $B$ has only a finite number of ordinary cusps $q_1, \ldots, q_t$ as singularities. These cusps are exactly the points over which $f$ is $totally$ $ramified$. Moreover $R$ is isomorphic to the normalization of $B$, in particular it is $smooth$.
 
Now we have

$f^*(B)=R + R'$,

where $R'$ is another irreducible curve, isomorphic to $R$, which meets $R$ in a finite number of points $p_1, \ldots, p_t$. Notice that $R'$ is $not$ a component of the ramification locus, since the latter consists of $R$ alone.


Moreover

 - $R$ and $R'$ are *tangent* at $p_1, \ldots, p_t$;
 - $p_1, \ldots ,p_t$ are the preimages of the cusps $q_1, \ldots, q_t$.

Summing up, in this case your $S$ is the set whose elements are the points $p_1, \ldots ,p_t$. They correspond to the points where the ramification divisor $R$ meets the curve $R'=f^*(B)   \setminus R$. In other words, they come from the singular points of the *branch* divisor $B$ (and not from the ramification divisor $R$, which is smooth). 

The point is that a general triple cover is not a Galois cover, so over the branch locus $B$ there are both points where $f$ is ramified (the curve $R$) and points where it is not (the curve $R'$). 

If you consider instead a Galois cover, say with group $G$, then every preimage of a branch point is a ramification point (and the stabilizers of points lying on the same fibre are conjugated in $G$). In this case there are formulae relating the ramification number of a point on $X$ with the ramification numbers of the components of the ramification locus passing through it. 

See Pardini's paper "Abelian covers of algebraic varieties" for more details.