Setting

In instanton gauge theory, given a $G$-principal bundle $P\to X^4$, the orientability of the moduli space of ASD connections $$\mathcal{M}_k = \{A \in L^{2}_{k}(X, \Lambda^1 \otimes\mathrm{ad}(P))\ | \ F_A^+ = 0\}/\mathcal{G}_{k+1}$$ (the subscript is a Sobolev parameter) is equivalent to the triviality of the determinant bundle of (the family of) operator(s) $$D_A : L^{2}_{k}(X, \Lambda^1 \otimes\mathrm{ad}(P)) \to L^{2}_{k-1}(X,(\Lambda^+ \oplus \Lambda^0)\otimes\mathrm{ad}(P))\\\\$$ $$D = d_A^+\oplus d_A^*.$$ This is a line bundle over the space of connections modulo gauge $\mathcal{B}_k$ (suppose we don't have any reducible for simplicity), $$\det D\to \mathcal{B}_k$$ where the fiber is given by $\det D_A := \Lambda^{\dim \ker D_A} \ker D_A \otimes \Lambda^{\dim \mathrm{coker} D_A} \mathrm{coker} D_A$

Suppose that $\mathcal{M}_k$ is smooth. We know that the tangent space at $[A] \in \mathcal{M}_k $, is given by $$T_{[A]}\mathcal{M}_k\simeq \ker d_A^+/ Im(d_A)$$ thus it seems to me that orientability should be equivalent to the bundle $$\bigsqcup_{[A] \in \mathcal{M}_k}\Lambda^{top}\ker d_A^+/ Im(d_A)\to \mathcal{M}_k$$ being trivial. This bundle is a priori different from $\det D$. Indeed, even if $\ker d_A^+/ Im(d_A) = \ker D_A$, in $\det D$ we are tensoring with the top power of the cokernel. What is the point of the cokernel tensor factor?

I know that we want $S_1 = \ker d_A|_{L_k^2(\Lambda^0\otimes \mathrm{ad}P)}$ and $S_2 = \mathrm{coker}\ d_A^+\subset L_k^2(\Lambda^+\otimes \mathrm{ad}P)$ to be trivial in order to have a smooth neighborhood at $[A]$. And that $\mathrm{coker}D_A > S_1 + S_2$ but maybe $\mathrm{coker}D_A$ is not trivial, so we don't have that $\det D_A = \Lambda^{top}\ker d_A^+/ Im(d_A) $ (maybe I am wrong).

Edit after M. Miller's answer In this case $\mathrm{coker}D_A = S_1 \oplus S_2 = \mathrm{coker} d_A^* \oplus \mathrm{coker} d_A^+$ (and consequently $[A] \in \mathcal{M}$ has a smooth neighbourhood iff $\mathrm{coker} D_A = (0)$). Indeed $$S_1 = \ker d_A|_{L_k^2(\Lambda^0\otimes \mathrm{ad}P)} \simeq \mathrm{coker} \ d_A^*.$$ Moreover
$$(\ker d_A^*)^\perp = Im \ d_A|_{L_k^2(\Lambda^0\otimes \mathrm{ad}P)} < \ker d_A^+$$ therefore $d_A^+ (\ker d_A^*) = Im \ d_A^+$ and consequently $$D_A (\ker d_A^*) = Im \ d_A^+ \quad \Rightarrow Im \ D_A = Im \ d_A^+ \oplus Im \ d_A^*$$