= Solution
Fix canonical single-particle order $(1\uparrow,1\downarrow,2\uparrow,2\downarrow)$ and write the occupation ket as $|n_1n_2n_3n_4\rangle$. The six states in the printed order are $|1010\rangle$, $|1001\rangle$, $|0110\rangle$, $|0101\rangle$, $|1100\rangle$, $|0011\rangle$. The last state is <double occupancy> on site 2.
For a <fermionic raising and lowering operator> acting on mode $j$, the sign is $(-1)^{\sum_{k<j}n_k}$, and annihilation requires that mode occupied while creation requires it empty. Starting from the second basis state, the hop that reaches the last state is
$$
c_{2\uparrow}^\dagger c_{1\uparrow}|1001\rangle
=c_{2\uparrow}^\dagger|0001\rangle=|0011\rangle.
$$
It therefore contributes \b[$H_{6,2}=-t$]. Starting from the third state instead,
$$
c_{2\downarrow}^\dagger c_{1\downarrow}|0110\rangle
=c_{2\downarrow}^\dagger|0010\rangle=-|0011\rangle,
$$
because the occupied $2\uparrow$ mode precedes $2\downarrow$. The <Hamiltonian>'s leading $-t$ makes \b[$H_{6,3}=+t$]. Finally $n_{2\uparrow}n_{2\downarrow}|0011\rangle=|0011\rangle$, so \b[$H_{6,6}=U$]. No other displayed basis state reaches $|0011\rangle$ in one allowed hop. These signs verify the three nonzero entries without assigning distinguishable-particle hopping signs to fermions.
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