Let $R$ be a commutative ring and $x$ is an $R$-regular element. It's a fact that the Koszul complex in this case is self-dual, i.e. $Hom_R(K^R(x),R)=K^R(x)[-1]$. This is true for a regular sequence too, but I'm sticking to the simple case.
My concern here is: the indices don't match. Look at the right-hand side. The index is defined to be $$(K^R(x)[-1])_i=(K^R(x))_{i-1}.$$ Since $K^R(x)$ is concentrated at $0$ and $1$, $K^R(x)[-1]$ is concentrated at $i$ such that $i-1=0,1$, i.e. $i=1,2$. Apparently, my argument is flawed somehow. In the solution for this problem, a note I'm reading claims that $K^R(x)[-1]$ is concentrated at $-1,0$.
What is my mistake?
This boils down to how you define $C[-1]$ (the issue is the definition changes if you're working with a homologically or a cohomologically graded complex, and sometimes we write $C[1]$ instead).
In your case, it is more or less clear that the dual of $K^R(x)$ is concentrated in degrees $-1$ and $0$, and $C[-1]$ on cochain complexes acts by $C[-1]^n = C^{n+1}$, contrary to what you write (and introduces a sign in the differential). This then gives what you want, i.e. you are desuspending $K^R(x)$.