= Solution
Write $\varepsilon=n_an^a\in\{-1,+1\}$. The <non-null hypersurface projection> is $P^a{}_b=\delta^a{}_b-\varepsilon n^an_b$, reproducing both signs in the question. It annihilates the normal and is the identity on tangent vectors to the <hypersurface>.
Represent $V$ by the tangent to a curve $\lambda$ through $p$. The curve $\phi\circ\lambda$ lies in $\Sigma$, so its tangent $d\phi_pV$ is tangent to $\Sigma$. Orthogonality therefore gives $n_a(\phi_*V)^a=0$, and
$$
\boxed{P^a{}_b(\phi_*V)^b=(\phi_*V)^a.}
$$
This uses the stated smooth <hypersurface> assumption; the dimension relation alone would not make the image of an arbitrary smooth map a regular hypersurface.
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