= Solution
Let $m=\dim M$, $n=\dim N$ and $k=\operatorname{codim}_N S$. 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 map>
$$
G=(x^1,\ldots,x^k)\circ f:f^{-1}(U)\longrightarrow\mathbb R^k.
$$
The derivative $L=d(x^1,\ldots,x^k)$ at a point of $S\cap U$ is surjective and has kernel $T S$, by the slice description. The printed sum-of-spaces condition gives
$$
L\bigl(df_r(T_rM)\bigr)=\mathbb R^k\qquad
\text{for every }r\in f^{-1}(S\cap U).
$$
Indeed 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>, $dG_r=L\circ df_r$ 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>. \b[Its dimension is $\boxed{m-k=m-n+\dim S}$]. Moreover,
$$
\boxed{T_p(f^{-1}(S))=\{v\in T_pM:df_p(v)\in T_{f(p)}S\}.}
$$
This follows either from the slice coordinates or from $T_pG^{-1}(0)=\ker dG_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.
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