Let $M^{n}$ be a compact, orientable, differentiable manifold with boundary $\partial M$ and $(\tilde{M}^{n+1},\tilde{g})$ be an oriented Riemannian manifold and let $x:M\longrightarrow\tilde{M}$ be an isometric immersion with $x(int M)\subset intB$, $x(\partial M)\subset \partial B$, $B$ being a smooth compact domain in $\tilde{M}$ that is diffeomorphic to a Euclidean ball. Let $X:(-\epsilon,\epsilon)\times M\longrightarrow\tilde{M}$ be a smooth variation such that $X(0,\cdot) = x$, $X|_{\partial M}\subset \partial B$, $X|_{intM}\subset intB$ for all $t$. Define the volume functional of $X$ to be $V(t)=\int_{[0,t]\times M} X^{*}d\tilde{M}$, where $X^{*}d\tilde{M}$ is the pullback volume form of $d\tilde{M}$ and let us write $Y = \partial_{t} X|_{t=0}$.
$\textbf{Claim}$: $V'(0) = \int_{M}\tilde{g}(Y,\eta)\,dM$, where $\eta$ is the unit normal field to $M$.
In the proof, the author fixes an arbitrary point $p\in M$ and lets $e_{1},\ldots,e_{n}, e_{n+1} = \eta$ be a positive oriented orthonormal frame around $x(p)$; $\eta$ is the uni normal field along $M$. Then, he writes, by definition of the pullback we have that $$\begin{align}X^{*}d\tilde{M} & = d\tilde{M}(\partial _{t}X,d_{(t,p)}X(e_{1}),\ldots,d_{(t,p)}X(e_{n}))\,(dt\wedge d\tilde{M})\\ & \overset{Q1}{=} \tilde{g}(Y,\eta_{t})\,(dt\wedge d\tilde{M}),\end{align}$$ where $\eta_{t}$ is the unit normal field of $X$. He then writes, that by a direct calculation one obtains $$\begin{align}\frac{d}{dt}V(0) & \overset{Q2}{=} \int_{M}\end{align}\tilde{g}(Y,\eta)\,dM.$$ Could someone please provide some details on how to obtain the equalites Q1 und Q2?