I am not exactly sure what your question means.
If I have a category $\mathcal{C}$, that I want to be enriched over a category $D$, but I don't suppose a priori that $\mathcal{D}$ has a monoidal structure.
Now to define the composition as a map in $\mathcal{D}$, what do I need?
First, I have to take two composable $Hom$ objects : $Hom(A,B)$ and $Hom(B,C)$, and I want a map that ends up in $Hom(A,C)$. I don't know yet what should be the source of this map, but I know that it has to be computed inside $\mathcal{D}$ from the two existing objects $Hom(A,B)$ and $Hom(B,C)$. So for now I will denote $Hom(B,C)\boxtimes Hom(A,B)$ this object, so that the composition of morphisms is actually a map $\circ: Hom(B,C)\boxtimes Hom(A,B) \to Hom(A,C)$ in $\mathcal{D}$. I will try to make as little assumptions as possible on the operation $\boxtimes$, while still making is a composition.
I want this operation to be functorial, so that I can use $circ : Hom(B,C)\boxtimes Hom(A,B) \to Hom(A,C)$ to define an arrow
$Hom(C,D)\boxtimes(Hom(B,C)\boxtimes Hom(A,B)) \to Hom(C,D) \boxtimes Hom(A,C)$. In general this means that my composition will respect the other computations that can happen inside $\mathcal{D}$, which is the whole point of enriched categories.
I know that I want the composition to be assocative, in other words, I want the two compositions $Hom(C,D)\boxtimes (Hom(B,C) \boxtimes Hom(A,B)) \to Hom(C,D)\boxtimes Hom(A,C) \to Hom(A,D)$
$(Hom(C,D)\boxtimes Hom(B,C))\boxtimes Hom(A,B) \to Hom(B,D)\boxtimes Hom(A,B) \to Hom(A,D)$ to be equal. But this doesn't a priori makes sense because their sources are different, so these two arrows are not comparable. the only way to make these arrows comparable is to assume that there is an invertible morphism $\alpha : Hom(C,D)\boxtimes (Hom(B,C) \boxtimes Hom(A,B)) \simeq (Hom(C,D)\boxtimes Hom(B,C)) \boxtimes Hom(A,B)$ in $\mathcal{D}$
Next I want each of the Hom(A,A) to have an identity. Given that my $Hom$ are not sets anymore, I can't talk about their elements, but I can use the trick of saying that an element is the same of a map from a distinguished object. I have to have a distinguished object $I$ in $\mathcal{D}$. Even by being as general as possible, and assuming nothing yet on $I$, I have a map $i : I \to Hom(A,A)$ in $\mathcal{D}$.
Now I also want the identity to be a left unit, that is considering
$I\boxtimes Hom(A,B) \to Hom(B,B) \boxtimes Hom(A,B) \to Hom(A,B)$
should be the identity of $Hom(A,B)$. But again this is not possible, since its source is $I\boxtimes Hom(A,B)$, and not just $Hom(A,B)$. So to express this property, I need to have a morphism $I\boxtimes Hom(A,B) \to Hom(A,B)$ in $\mathcal{D}$.
You could do the same thing for the right unit, and you finally get that $\boxtimes$ has to satisfy all the axioms of a monoidal structure on the category $\mathcal{D}$. So in the end, if you want to enrich the category $\mathcal{C}$ over the category $\mathcal{D}$, in order to express all the properties you want the composition to satisfy, you need $\mathcal{D}$ to be a monoidal category, and so this really is the context in which you want to work.
I am not really sure what you call linearity in this context, given that there is a priori no addition, nor scalar multiplication of the objects of the category $\mathcal{D}$, and monoidal categories have a definition more general than the tensor product on the category of modules.
