= Solution
Expand the field operator in orthonormal localized orbitals, $\hat\psi(\mathbf r)=\sum_i\phi_i(\mathbf r)c_i$. For a one-electron continuum <Hamiltonian> $h_0=\hat p^2/(2m^*)+V(\mathbf r)$, projection gives
$$
H_0=\sum_{ij}h_{ij}c_i^\dagger c_j,\qquad
h_{12}=\int d^3r\,\phi_1^*(\mathbf r)
\left[\frac{\hat p^2}{2m^*}+V(\mathbf r)\right]\phi_2(\mathbf r).
$$
The coefficient in the displayed <lattice> <Hamiltonian> is $-t_{12}$. Thus the <projected hopping matrix element and sign convention> require
$$
\boxed{-t_{12}=h_{12}=K_{12}+V_{12},\qquad
K_{12}=\int d^3r\,\phi_1^*\frac{\hat p^2}{2m^*}\phi_2.}
$$
<Orthogonality> states $\int\phi_1^*\phi_2=0$; it does not imply $\int\phi_1^*V\phi_2=0$ for a spatially varying potential. If the off-diagonal potential is assumed zero, for example when it is a constant, the hopping coefficient is $h_{12}=K_{12}$. This is the kinetic integral requested by the paper, but it equals $-t_{12}$ under the paper's own leading-minus convention. A convention writing the hopping term with a plus sign would instead call $h_{12}$ itself $t_{12}$.
\b[The printed positive equality is not a consequence of the stated <orthogonality> assumption and the displayed sign convention.] <Orthogonality does not diagonalize a varying potential> gives an explicit counterexample to the underlying inference. The general projected relation above states the supported result; setting $V_{12}=0$ still leaves $-t_{12}=K_{12}$ in the question's convention. Rephasing an orbital changes both $h_{12}$ and the coefficient of the corresponding operator together; it does not remove this algebraic sign mismatch in a fixed convention. The spectrum in the remaining two-site calculation depends on $|t|^2$, so those parts are unaffected.
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