This is a partial answer, but I hope it's at least helpful.
I also can't see how to reconcile the statements
(1) $K c_{ks} = -i\sigma^y_{ss'} c^\dagger_{-k, s'}$ and
(2) $ K \hat{c}_{\uparrow} K^{-1} = \hat{c}_{\downarrow}; K \hat{c}_{\downarrow} K^{-1} = -\hat{c}_{\uparrow}$.
When working with antilinear objects, we first need to establish a phase convention. Suppose you have two bases for the 1D complex vector space formed by the vacuum - $|0\rangle$, and $|0'\rangle = e^{i\phi} |0\rangle$. How does $C$ act on these bases? If you suppose that $C | 0 \rangle = | 0 \rangle$, then you end up with the strange-looking
$$ C | 0' \rangle = e^{-i\phi} |0\rangle = e^{2i\phi} |0'\rangle.$$
where $C$ is complex conjugation. This is often suppressed by picking a phase convention for the vacuum that ensures that $C |0\rangle = |0 \rangle$. Choosing the same convention for all basis kets, defined as $c^\dagger_{s} |0 \rangle$ means that $Cc_{s} C \equiv c_s$.
In short, the action of time reversal can change the phases in front of operators, but does not mix orthogonal states.
However, I believe that (2) is consistent with the claimed superconducting order parameter transformation.
The SC order parameter operator (as a matrix) is written such that
$$ \hat{\Delta}(k) = \begin{pmatrix} \hat{c}_{-k\uparrow} & \hat{c}_{-k\downarrow} \end{pmatrix}\begin{pmatrix} \Delta_{\uparrow\uparrow}(k) & \Delta_{\uparrow\downarrow}(k) \\ \Delta_{\downarrow\uparrow}(k) & \Delta_{\downarrow\downarrow}(k) \end{pmatrix}
\begin{pmatrix} \hat{c}_{k\uparrow} \\ \hat{c}_{k\downarrow} \end{pmatrix}
= \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'} $$
In second quantisation, symmetries acts on operators by the conjugation action (i.e. in the adjoint representation). Due to decades of sloppy notation, we end up with garbage like
$$ \tilde{K} \hat{\Delta}(k) = K \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'}K^{-1} $$
$$ = -i \sigma^y C \Delta_{ss'}(k)\hat{c}_{-k s}\hat{c}_{k s'} i \sigma^y C
\hspace{3em} \text{(*)}
$$
Now there is a problem - we defined how $C$ acts on $|\uparrow\rangle$ for a single spin, but how does it act in the full Hilbert space, $\text{Span}\{ c^\dagger_{r,s} |0\rangle ,r \in \text{Points in space} \}$ ?
Suppose that we make the choice of phase $C c^\dagger_{r, s}|0\rangle = c^\dagger_{r, s}|0\rangle$. Then complex conjugation acts non-trivially on $k$ space eigenvectors -
$$C c^\dagger_{k,s}|0\rangle = C\sum_{r}e^{-ik\cdot r} c^\dagger _{r, s} |0\rangle$$
$$ = \sum_{r}e^{ik\cdot r} C c^\dagger _{r, s} |0\rangle = \sum_{r}e^{ik\cdot r} c^\dagger _{r, s} |0\rangle = c^\dagger_{-k,s} |0\rangle$$
Using the anticommutators, this argument shows that in the standard phase convention, $C c_{k,s} C = c_{-k, s}$.
Using this fact gets you what you want.
$$ K\hat{\Delta}(k)K = +\sigma^y \Delta_{ss'}^*(k) \hat{c}_{k, s} \hat{c}_{-k, s'} \sigma^y $$
$$= -\sigma^y \Delta_{ss'}^*(-k) \hat{c}_{-k, s'} \hat{c}_{k, s}\sigma^y $$
which is the claimed result, since $\Delta(k) = -\Delta(-k)^T$.