 Solution

Let $m=\dim M$, $n=\dim N$ and $k={\mathrm{codim}}_{N}S$. Fix $p\in f^{-1}(S)$ and choose the allowed slice chart for an embedded submanifold on a neighborhood $U$ of $f(p)$, so that $S\cap U$ is given by $x^1=\cdots =x^k=0$. On $f^{-1}(U)$ define the smooth mapThe derivative $L=d(x^1,\dots ,x^k)$ at a point of $S\cap U$ is surjective and has kernel $TS$, by the slice description. The printed sum-of-spaces condition givesIndeed any target tangent vector is a sum of an image vector and a tangent vector to $S$, and $L$ kills the latter. By the chain rule, $d{G}_r=L\circ d{f}_r$ is therefore surjective. This condition is precisely transversality of a map to a submanifold; the original sum need not be direct.

We have $G^{-1}(0)=f^{-1}(S)\cap f^{-1}(U)$. Part (b), applied to $G$, gives slice charts for an embedded submanifold of codimension $k$ around every point of this zero set. Since the construction is available around every $p\in f^{-1}(S)$, the charts give its global structure as an embedded submanifold with the subspace topology. Its dimension is $\boxed{m-k=m-n+\dim S}$. Moreover,This follows either from the slice coordinates or from $T_p{G}^{-1}(0)=\ker d{G}_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.