Let me expand jvp's answer, giving a picture of the situation in the case of a $general$ flat triple cover $f \colon X \to Y$.
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 $smooth$.
One has the equality of divisors
$f^*(B)=2R + R'$,
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.
- $R$ 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 $B$ (whereas the ramification divisor $R$ is smooth).
This is easy to see; a good reference is Miranda's paper "Triple covers in algebraic geometry".
Anyway, the crucial fact here 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 any 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.