Definition $4.18$ in O'Neill's book Semi-Riemannian geometry, with Applications to Relativity states that given a unit vector field $U$ normal to a hypersurface $M\subset \overline M$, then, there is a $(1,1)$ tensor field $S$ on $M$ such that $$ \langle S(V), W\rangle = \langle II(V,W), U\rangle \ \, \forall V,W \in \mathfrak X(M) , $$ called the shape operator of $M\subset \overline M$ derived from $U$.
Just after this definition it is also stated (Lemma $4.19$) that, $S(V)=-\overline D_VU$, because $$ \langle S(V), W\rangle = \langle II(V,W), U\rangle = \langle \overline D_VW,U\rangle =- \langle \overline D_VU, W\rangle \ , \ \forall W\in \mathfrak X(M). $$ Then, I have two (related?) questions:
(1) Why is $S$ a tensor field on $M$? I mean, in general, $\overline D_VU$ would have a tangent component to $M$ (i.e. ${\rm tan}\overline D_VU \in \mathfrak X(M)$) but also a normal component (with ${\rm nor}\overline D_VU \notin \mathfrak X(M)$), so it is a vector field of the tangent space of the ambient manifold $\overline M$, not of the tangent space of $M$, isn't it? Or, is there any evident way to see that actually, $\overline D_VU = {\rm tan}\overline D_VU\in \mathfrak X(M)$?
(2) From the second of the above equations, we also have $$ \langle S(V), W\rangle = -\langle \overline D_VU, W\rangle = -\langle {\rm tan}\overline D_VU + {\rm nor}\overline D_VU, W\rangle= -\langle {\rm tan}\overline D_VU, W\rangle \ , \ \forall W\in \mathfrak X(M) $$ why is then $S(V)$ defined as $-\overline D_VU$ and not as $\ -{\rm tan}\overline D_VU$? is it, as in the previous question, because $\overline D_VU = {\rm tan}\overline D_VU$?