Yes, one can always do this. Start with a $p$-centered local coordinate system $(x^\sigma,y^\rho)$ on an open $p$-neighborhood $U\subset M$ such that $S\cap U$ is given by $y^\rho=0$ $(1\le\rho\le r)$. Then there will exist functions $F^\sigma_\rho(x)$ such that $V$ along $S\cap U$ is given by the equations $$ \mathrm{d}x^\sigma + F^\sigma_\rho(x)\,\mathrm{d}y^\rho = 0 $$ (summation convention assumed). This is the same as $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) - y^\rho\,\mathrm{d}(F^\sigma_\rho(x)) = 0. $$ Since $y^\rho=0$ along $S$, this says that $V$ along $S$ is defined by $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) = 0, $$ so take $\bar x^\sigma = x^\sigma+F^\sigma_\rho(x)y^\rho$, and let $(\bar x^\sigma,y^\rho)$ be your new coordinate system. (One can easily check that these are local coordinates in an open neighborhood of $S\cap U\subset U$.)
Added Remark: Just a comment on your terminology: Literally, what you are defining is not a splitting of the normal bundle of $S$, but rather a choice of a normal bundle along $S$, and you are asking whether there are any local invariants that distinguish such choices. As the above shows, the answer is 'no'. In fact, there is even a global version: For two normal bundles $V_1$ and $V_2$ along a closed submanifold $S$, there is a diffeomorphism of $M$ that fixes $S$ and carries $V_1$ to $V_2$.
Note also that, in the holomorphic category (in contrast to the smooth case), normal bundles need not be unique in this sense, and may not even exist.