Let me use the transformation $\overline g = e^{2f}g$ to simplify some notations (and I guess your formula also use this convention). Near a point $p\in N,$ let $\{e_i\}$ be an orthonormal frame with respect to $g,$ and $\eta$ be a normal. Then with respect to $\overline g,$ we have $\overline e_i = e^{-f}e_i$ form an orthonormal frame near $p$ and $\overline\eta=e^{-f}\eta$ being the normal. Then
\begin{align}
\overline h_{ij}
& = \langle\overline\nabla_{\overline e_i}\overline e_j,\overline\eta \rangle_{\overline g}\\
& = e^{2f}\langle e^{-2f}\overline\nabla_{e_i}e_j, e^{-f}\eta\rangle_g\\
& = e^{-f}\langle \nabla_{e_i}e_j-\delta_{ij}\nabla f,\eta \rangle_g\\
& = e^{-f}( h_{ij} -\langle \nabla f,\eta\rangle_g\delta_{ij})\\
\end{align}
where we use the transformation of the Levi-Civita connection, i.e.,
$$\overline\nabla_X Y = \nabla_XY + (Xf)Y + (Yf)X - \langle X,Y\rangle_g\nabla f.$$
Thus
\begin{align}
\overline H
& = \sum_{i=1}^{n-1}\overline h_{ii}\\
& = \sum_{i=1}^{n-1}e^{-f}(h_{ii} - \langle \nabla f,\eta\rangle_g)\\
& = e^{-f} (H-(n-1) \langle \nabla f,\eta\rangle_g),
\end{align}
where $n$ is the dimension of $N,$ so in your case $n-1=2.$

tangentvector field to $M$ rather than the normal vector field? $\endgroup$