Let $M^n \subset \mathbb{R}^{n+1}$ oriented submanifold with the induced euclidean metric. For $t \in \mathbb{R}$, let $\psi_t : M \to \mathbb{R}^{n+1}$, $p \mapsto p + tN(p)$, where $N$ is the unit normal vector field over $M$. For small values of $t$, $\psi_t$ is an inmersion and $(d\psi_t)(T_pM) = T_pM$. In fact,
$$ (d\psi_t)_p(v) = v + t dN_p(v), \quad \forall v \in T_pM. $$
Let $g_t$ the metric over $M$ obtained from pullback of the scalar product in $\mathbb{R}^{n+1}$, that is:
$$ g_t(v,w) = \langle (d\psi_t)_p(v),(d\psi_t)_p(w)\rangle. $$
Since $d\psi_t$ is a vector space isomorphism, $N_t = N$. This does not mean that the Weingarten $A_t$ endomorphism of $(M,g_t)$ does not depend of $t$. My teacher said that
$$ (II_t)(v,w) = -g_t((dN_t)_p(v),w) \stackrel{(*)}{=} -\langle (dN_t)_p(v), (d\psi_t)_p(w)\rangle $$
I can't see why $(*)$ is true. I thought that might be
$$ -\langle (d\psi_t)_p((dN_t)_p(v)), (d\psi_t)_p(w)\rangle $$
Finally we must obtain
$$ (II_t)_p(v,w) = g_{p}\left(A_{p} v, w-t A_{p} w\right) $$
Edited:
The second fundamental form $II_t$ of $(M,g_t)$ is
$$ \begin{split} (II_t)_p(e_i,e_j) = k_i(p)\left(1 - tk_j(p)\right) \delta_{ij}. \end{split} $$
where $\{e_i\}$ is a ortogonal basis of principal directions of $(M,g)$. This base is not $g_t$-ortonormal, but $$ \left\{\frac{1}{1-t k_{1}(p)} e_{1},\dots, \frac{1}{1-t k_{n}(p)} e_{n},\right\} $$ it is and $II_t$ diagonalize. All of this is strictly directed with the tube formula of compact hypersurfaces of euclidean space $$ \int_{\Omega_r} f d v_{g_{0}}=\int_{p \in M}\left(\int_{0}^{r} f\left(p+t N_{p}\right) \prod_{i=1}^{n}\left(1-t k_{i}(p)\right) d t\right) d v_{g} $$ where $\Omega_r$ is the interior half tube of $M$, i.e, the image of $F$. In particular, for $f \equiv 1$
$$ vol(\Omega_r)= \sum_{i = 0}^{n}\binom{n}{i}\dfrac{r^{i+1}}{i+1}\int_{M}\sigma_i(k_1,\dots,k_n) \ d vol_M, $$
(Weyl tube formula or Steiner formula)