It's pure group theory. In the normal electroweak theory, the Yukawa interaction (which is what always produces the fermionic mass terms after the symmetry breaking - after the Higgs gets a vev) is simply (schematically)
$$ h_2 \cdot L_{L2} \cdot L_{R1} $$
where the Higgs $h_2$ and the left-handed leptons $L_L$ are doublets (contracted with one another) and the right-handed lepton is a singlet.
You want a more complex theory where the Higgs is a triplet of the weak $SU(2)$ group. The leptons are at most doublets, so the only way how you can get a scalar (Lagrangian) is to contract the Higgs triplet with two doublets.
That's why you need two copies of the left-handed fermions (not one left plus one right – the right-handed guys are singlets) and the resulting mass term is a Majorana mass term (not a Dirac mass term). The group theory of the contraction of a triplet, doublet, doublet is equivalent to
$$ S^\dagger \cdot (\vec V_3\cdot \vec \sigma) \cdot R $$
So it's a $2\times 2$ matrix created from the triplet/vector $\vec V_3$ sandwiched between two doublets. Note that $\vec V_3\cdot \vec \sigma$ with the Pauli matrices vector is just a way to write a triplet as a $2\times 2$ matrix, the only way.
In your formula, $f_{\alpha\beta}$ are just the coefficients and the $i\sigma^2$ and $C^{-1}$ stuff is needed for the removal of the complex conjugation. You know, the triplet of the Pauli matrices naturally gives us matrices i.e. a part of the representation $2\otimes {\bar 2}$. However, we want it in the basis of $2\otimes 2$ without a conjugation. $2$ is isomorphic to $\bar 2$ but to get the components right, we always need to play with $\sigma^2$.
It's convenient to rewrite the triplet as a $2\times 2$ matrix because the Yukawa term (which gives rise to the mass term for fermions after a Higgs vev) may be written as a nice matrix element of the style $\vec u \cdot M \cdot \vec v$ that physicists know very well. If you didn't use the matrix, you would have to use the coefficients
$$ K_{a,b,c} $$
where $a=1,2,3$ while $b,c=1,2$ and remember lots of the values in this tensor. However, with the matrix rewriting, we can work with well-known matrix products supplemented by $C^{-1}$ and $\sigma$.