The point
An abelian model category is that when a category which is abelian and has a model structure, along with a compatibility condition between those two things. So in particular, you need the cofibrations to be monomorphisms with cofibrant cokernel and the fibrations to be epimorphisms with fibrant kernel. When you have an abelian model category (i.e. abelian category with a model structure and a compatibility condition) then you get a two cotorsion pair pairs from the classes of fibrant objects and acyclic cofibrant objectsit. Conversely, given a cotorsion pair you can construct a model structure if your cotorsion pair is nice (it has to be "complete" which means you need to have enough projectives and enough injectives).
In your case, $R$-mod has two complete cotorsion pairs: $(C, W\cap F)$ and $(C\cap W, F)$, by Prop 2.2. in the linked paper. You can promote those to pairs on $I^{R-mod}$ via $(I^C, I^{W\cap F})$ and $(I^{C\cap W}, I^F)$ using the fact that Ext groups between diagrams should be computed object-wise (I can't imagine how else you would compute them). It's also easy to see that a diagram of elements of the form $W\cap F$ is an intersection of diagrams from $I^W$ and $I^F$, so you can write $I^{W\cap F}$ as $I^W \cap I^F$. You say in your question that the diagrams have enough projectives and injectives, so this means you get complete cotorsion pairs, . All you have to do to use Theorem 2.5 of the paper and hence get an abelian model structure is to show $I^W$ is thick, by Theorem 2.5 of the linked paperi.e. Note that the cofibrations have to be monomorphisms with cofibrant cokernel closed under retracts and satisfies the fibrations have to be epimorphisms with fibrant kernel2-out-of-3 property. That's what it means to Both should be an abelian model categoryeasy, since $W$ is thick.

