Kashiwara's Theorem Proof for Right D-modules

I'm reading a proof of Kashiwara's Theorem where the authors are basing everything off right D-modules, and there seems to be a sign error somewhere that I can't figure out.

To set notation, let $i:X\to Y$ be the closed embedding, with $X:=(f=0)$. Set $\partial$ to be such that $[\partial,f]=1$, and set $s:=f\partial$. Let $i_0^!:=H^0(i^!)$, with $i^!(N):= \text{RHom}_{D_Y}(D_{X\to Y},N)$ and $i_*(M):= M\otimes_{D_X}D_{X\to Y}$. Explicitly, M will denote right $D_X$ and N, right $D_Y$ modules; N here is supported on $X$ of course.

The first half of the concern is the following: I understand that, in this case, $D_{X\to Y}\cong D_X[\partial]$, and so $i_*(M)\cong M[\partial]$. They then, however, claim that $M\partial^k$ is the $(-k)$-eigenspace of $\cdot s$. With the usual convention of right actions by derivations having a minus sign, I agree with this. For example,

$(m\partial)\cdot s=-m(\partial s)=-m(s\partial+\partial)=-m\partial$ since $m\cdot s=(m\cdot f)\cdot \partial=0$ as $f=0$ on $X$, and so its action on a $D_X$-module is 0. Moreover, $\partial:M\partial^k\to M\partial^{k+1}$ and $f:M\partial^k\to M\partial^{k-1}$

The other issue is showing essential surjectivity of $i_*$. If $N$ is supported on $X$, then we again try to define the $0,-1,-2,...$ eigenspaces. However, if for example, $n\cdot s=0$m then $(n\cdot\partial)\cdot s=n\cdot(\partial s)=n\cdot(s\partial+\partial)=n\cdot\partial$, so $n\cdot\partial$ has eigenvalue $1$; the 1st case has $\partial$ decreasing the eigenvalue, and the 2nd has it increasing.

Where does the sign issue lie exactly? Thanks in advance! <3