In picture below,$\iota$ is inclusion.I guess $v$ is outer normal vector. Why the volume form is a vector times $dx$ ? I mean that the Riemannian volume form is $\sqrt{det(g)}dx_1...dx_n$ , is a function times $dx$.But in the picture below ,it is a vector times $dx$.
Sorry for my poor English,I really don't know whether I have made myself clear.
In fact, I guess that $\iota_{\nu}$ is the interior multiplication by $\nu$, i.e. $$ \iota_{\nu} : \Omega^n(M) \rightarrow \Omega^{n-1}(M), $$ where $\iota_{\nu}(\omega)$ is the $(n-1)$-form on $M$ defined by $$ \iota_{\nu}(\omega)_p(v_1,\ldots,v_{n-1}) = \omega_p(\nu,v_1,\ldots,v_{n-1}). $$ So in your case, $d\mu_g$ is an $n$-form on $M$ and thus $\iota_{\nu}d\mu_g$ is an $(n-1)$-form on $M$. It remains to "pullback" it in $\partial M$, i.e. if you have the inclusion $i : \partial M \hookrightarrow M$, then $i^{\ast}\left(\iota_{\nu}d\mu_g\right) \in \Omega^{n-1}(\partial M)$ is the volume form (also written $\iota_{\nu}\left.d\mu_g \right|_{\partial M}$) of the boundary $\partial M$.