= Solution
Use signature $(-,+,+,+)$ and write $h_{ij}$ for the spatial <induced metric>, $D_i$ for its <spatial covariant derivative>, and $q^2=h^{ij}D_i\phi D_j\phi$. The negative <shift vector> convention gives $n^\mu=N^{-1}(1,N^i)$ and $\Pi=n^\mu\partial_\mu\phi$. Decomposing the <scalar field> derivative gives
$$
\partial_\mu\phi=-n_\mu\Pi+D_\mu\phi,\qquad g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi=-\Pi^2+q^2.
$$
Varying the <scalar field> matter <action> with respect to the inverse <metric tensor>, including the variation of $\sqrt{-g}$, gives the <stress-energy tensor>
$$
\boxed{T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}\left[\tfrac12g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi+V\right].}
$$
Raising both indices gives the requested $T^{\mu\nu}$. Contracting with the <unit normal> twice, or once with the <spatial projection tensor>, gives
$$
\boxed{\rho=\tfrac12\Pi^2+\tfrac12q^2+V=N^2T^{00},\qquad \mathcal J_i=-\Pi D_i\phi.}
$$
Projecting both indices of the <stress-energy tensor> onto the spatial hypersurface gives
$$
S_{ij}=D_i\phi D_j\phi+h_{ij}\left(\tfrac12\Pi^2-\tfrac12q^2-V\right),\qquad
\boxed{S=\tfrac32\Pi^2-\tfrac12q^2-3V,\quad \widetilde S_{ij}=D_i\phi D_j\phi-\tfrac13h_{ij}q^2.}
$$
\b[Two printed identifications require correction.] The <canonical momentum> of the displayed density is $\pi_\phi=\sqrt h\,\Pi$, not $\Pi$: explicitly $\mathcal L_m=N\sqrt h[\Pi^2/2-q^2/2-V]$ and differentiation with respect to $\dot\phi$ supplies $\sqrt h$. Thus $\Pi$ is the <normal scalar-field momentum>. Also, the printed trace-free stress has an extra factor $1/2$ and uses $\delta_{ij}$ where the general spatial <induced metric> requires $h_{ij}$. In an orthonormal spatial frame the latter becomes $\delta_{ij}$, but the coefficient is still one. For a gradient $(q,0,0)$ in that frame, direct projection gives $\widetilde S_{11}=2q^2/3$, rather than the printed $q^2/3$. The PDF's energy-density gradient is spatial $\partial_i\phi\partial^i\phi$; the TeX aid's spacetime index there is an OCR defect.
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