Let me attempt to answer your question as I understand it.
Let $x\in H^1(M)$ be the Poincaré dual of $[c]\in H_1(M)$ (all (co)homology groups are with $\mathbb{Z}_2$ coefficients). The result of Thom you state is that $w_1(M)\in H^1(M)$ is the Poincaré dual of $[S(f)]\in H_1(M)$.
It is well known that the cup product in cohomology coincides under Poincaré duality with the intersection of transverse representatives of homology classes. It follows that the parity of the number of intersection points of $S(f)$ with $c$ equals the Kronecker product $\langle w_1(M)\cup x,[M]\rangle\in\mathbb{Z}_2$.
Now use the identity $\langle a\cup b,c \rangle = \langle a, b \cap c\rangle$ relating cup, cap and Kronecker products (see any good Algebraic Topology book, eg Switzer) to conclude that $$\langle w_1(M)\cup x,[M]\rangle = \langle w_1(M),x\cap [M]\rangle = \langle w_1(M),c\rangle$$ as required.