There are two steps to working smoothly with sign conventions. The first is to do as much as possible categorically, where all choices of sign conventions are relegated to single black box. The second step is only to unpack the black box when necessary, usually at the very end of a calculation.

Let $\mathcal C$ be a symmetric monoidal category. I denote the monoidal structure by $\otimes$, the monoidal unit by $\mathbf 1$, and the symmetry by $\operatorname{flip}$. I assume moreover that $\mathcal C$ is enriched in abelian groups (maybe because it is enriched in $k$-modules for some commutative ring $k$). I assume also that $\mathcal C$ has kernels.

A *commutative algebra* in $\mathcal C$ is an object $A\in\mathcal C$ along with maps $m: A \otimes A \to A$ and $1: \mathbf 1 \to A$, such that: (commutativity) $m = m\circ \operatorname{flip}$, (associativity) $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$, and (units) $m \circ (1 \otimes \operatorname{id}) = \operatorname{id}$. A *(left) module* for $A$ is an object $M \in \mathcal C$ and a map $m : A\otimes M \to M$ such that $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$. Note that every left module is also a right module and in fact a bimodule by composing judiciously with $\operatorname{flip}$.

A morphism $\delta : A \to M$ is a *derivation* if $\delta \circ m = m\circ (\delta \otimes \operatorname{id}) + m\circ (\operatorname{id}\otimes \delta)$. Then the set $\operatorname{der}(A,M)$ of derivations is naturally an abelian subgroup of $\hom(A,B)$.

**Aside:** Suppose that $(\mathcal C,\otimes)$ is *closed* in the sense that for any object $X$, the functor $\otimes X : \mathcal C \to \mathcal C$ has a right adjoint $\underline{\hom}(X,-)$. It follows that $\underline{\hom}(-,-)$ is contravariant in the first component. Then in particular there is an object $\underline{\hom}(A,M) \in \mathcal C$. It has a subobject $\underline{\operatorname{der}}(A,M) = \ker\bigl( \underline{\hom}(A,M) \rightrightarrows \underline{\hom}(A\otimes A,M)\bigr)$. One of these maps corresponds to $m : A\otimes A \to A$ in the first component. The second corresponds to $m : A\otimes M \to M$, and uses the adjunctions a few times in its definition.

Anyway, I believe that your claim that $I = \ker(m : A\otimes A \to A)$ comes equipped with a canonical derivation is satisfied in any symmetric monoidal linear category with kernels. At worst, you need to include some more adjectives like "abelian" or "presentable". But it is checkable without reaching inside to see how the objects are builts

Let me describe now a notation for doing such computations. For any category $\mathcal C$ and any object $A\in \mathcal C$, a *generalized element* $x\in A$ is a morphism $X \to A$ for some $X\in \mathcal C$. The generalized elements with domain $X$ are the *$X$-points of $A$*. The Yoneda lemma implies that an object is determined by its generalized elements. (In fact, in a sense $A$ is determined by its canonical $A$-point $\operatorname{id}:A \to A$.) If $\mathcal C$ is symmetric monoidal, *global element* of $A \in \mathcal C$ is a morphism $\mathbf 1 \to A$. Note that global elements do not determine the object. Generalized elements in a symmetric monoidal category enriched over abelian groups can be manipulated like regular elements of regular abelian groups. The requirement is that in any expression, there are no duplications or deletions: the expression should be homogeneous linear in each element.

For example, suppose that you have generalized elements $x: X \to A$ and $y : Y \to A$. Then $xy = m(x\otimes y)$ is a generalized element with domain $X\otimes Y$. On the other hand, $yx \in A$ has domain $Y \otimes X$. There is a unique canonical isomorphism $\operatorname{flip}:X\otimes Y \to Y \otimes X$. Then an equation like "$\forall x,y\in A$, $xy = yx$" has the interpretation: "for all pairs of generalized elements $x : X \to A$ and $y : Y \to A$, we have an equality of morphisms $m\circ (x\otimes y) = m\circ (y\otimes x)\circ \operatorname{flip} : X\otimes Y \to A$." With this convention, you will never see signs. For example, a morphism $\delta : A\to M$ is a *derivation* if for all generalized elements $a,b \in A$, we have $\delta(ab) = a\delta(b) + b\delta(a)$.

This is all well and good, but how do you unpack it? The category of graded vector spaces is a monoidal category using the "convolution product" — this is what you're used to. Its symmetric structure is determined by the requirement that $\operatorname{flip} : [1] \otimes [1] \to [1] \otimes [1]$ is *minus* the identity, where $[1]$ denotes the one-dimensional vector space in degree $1$ (or in degree $-1$, whichever is your convention); it and its inverse $\otimes$-generate the category of graded vector spaces. You can skeletalize your category to decide that every object is a direct sum of various $[n] = [1]\otimes \dots \otimes [1]$. With respect to this skeletalization, $[m+n] = [m]\otimes [n] \overset{\operatorname{flip}}\to [n]\otimes[m] = [n+m]$ is the map $(-1)^{mn}$, which is the usual Koszul rule. This category has a distinguished Lie algebra, which is the abelian Lie algebra on the object $[1]$. (By "abelian", I mean that the Lie bracket is $0$.) The category of dg vector spaces is the category of representations of this Lie algebra (the Lie algebra action records the differential). It is a symmetric monoidal category in the same way that the representation theory of any Lie algebra is a symmetric monoidal category. In particular, forgetting from a dg vector space to a graded vector space is faithful and preserves the symmetric monoidal structure. So a morphism in dgVect is a morphism in gVect that is equivariant for the Lie algebra action.

It's then important to remember that forgetting from a graded vector space to a regular vector space is monoidal, but does not preserve the symmetric structure. So when working with generalized elements in gVect, you can say that a derivation should satisfy $\delta(ab) = a\delta(b) + b\delta(a)$. But when working with elements of the underlying vector space, you must remember that there was an implicit $\operatorname{flip}$ when the $a$ and $b$ switched places; for that (worse) convention, we have $\delta(ab) = a\delta(b) + (-1)^{|a||b|}b\delta(a)$.

Finally, you asked about derivations of non-zero degree. As in the aside, suppose that $\mathcal C$ is closed. Then one always has $\hom(\mathbf 1,\underline{\hom}(A,M)) = \hom(A,M)$ as abelian groups, and also $\hom(\mathbf 1,\underline{\operatorname{der}}(A,M)) = \operatorname{der}(A,M)$. When $\mathcal C$ is the category of graded or dg vector spaces, the monoidal unit is $\mathbf 1 = [0] = $ the one-dimensional vector space in degree $0$. Up to a grading convention, a *degree-$k$ differential* is a morphism $[k] \to \underline{\operatorname{der}}(A,M)$. It is not hard to unpack this definition: it says that $\delta(ab) = (-1)^{|a|k}a\delta(b) + (-1)^{|b|(|a|+k)}b\delta(a)$. I think one of your formulas must have been about degree-odd derivations.

It is an unfortunate but common occurrence for writers to use worse conventions. For example, remember that the canonical isomorphism $[k] \otimes X \cong X\otimes [k]$ is not the identity on underlying vector spaces. In the conventions above, the adjunction implies that a map $[k] \to \underline{\operatorname{der}}(A,M)$ is a map $[k] \otimes A \to M$ satisfying some condition. It is the same data as a map $A[k] = A\otimes [k] \to M$, but on underlying vector spaces those two maps are different.

The moral of the story is that you should never cut-and-paste sign-full formulas from anyone else. Rather, pick a convention and stick to it, and as much as possible make no choices at all by using a categorical framework.