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^fe_i$$\overline e_i = e^{-f}e_i$ form an orthonormal frame near $p$ and $\overline\eta=e^f\eta$$\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} - e^f\langle \nabla f,\eta\rangle_g\delta_{ij}\\ \end{align}\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 & = \overline g^{ij}\overline h_{ij}\\ & = e^{-2f}g^{ij}(e^f h_{ij} - e^f\langle \nabla f,\eta\rangle_g\delta_{ij})\\ & = e^{-f} (H-(n-1) \langle \nabla f,\eta\rangle_g), \end{align}\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 $n-1=2.$
