**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
$$D_A (\ker d_A^*) = Im \ d_A^+ \quad \Rightarrow Im \ D_A = Im \ d_A^+ \oplus Im \ d_A^*$$