Structure of triangulated category generated by category $B-$mod, where $B=A/J$

Let $$A$$ be a finite dimensional algebra over a field $$k,$$ $$J\subset A$$ some ideal of $$A$$ and $$B=A/J$$ be a quotient algebra. Assume that $$\text{Ext}^n_A(B_A,B_A)=0$$ for $$n>0$$ and $$B$$ has finite projective dimension as a right $$A-$$module. It is said in some paper that it is easy to see that the category $$B-$$mod generates the strict, triangulated subcategory $$\mathcal{D}$$ of $$D^b(A-\text{mod})$$ consisting of objects which have cohomology in the subcategory $$B-$$mod of $$A-$$mod and also that $$D^b(B-\text{mod})\rightarrow D^b(A-\text{mod})$$ factors through $$\mathcal{D}.$$

I am new in the ground of triangulated categories -- I understand well definitions, notions but I don't have certain intuitions.

Of course, category $$B-$$mod is abelian, but not triangulated. It seems for me informally that triangulated category generated by $$B-$$mod consists of objects which have cohomology in the subcategory $$B-$$mod but I would like to ask you a method, a way to formalize this intuition -- why it is true in a formal way?

I don't see also this factorization through $$\mathcal{D}$$ -- how we map an arrows of $$D^b(B-\text{mod})$$ to an arrows of $$\mathcal{D}?$$

Let me try to give a answer I hope will help, but this is not a complete answer though. I assume all the categories are bounded. I will assume that $$B-mod$$ is thick (maybe this is implied by the condition you give ?).
Denote $$T\subset D(A-mod)$$ the strict triangulated subcategory of $$D(A-mod)$$ generated by $$B-mod$$ and $$D\subset D(A-mod)$$ the subcategory of object with cohomology in $$B-mod$$.
First, $$T\subset D$$: if $$x \to y \to z$$ is a distinguished triangle, using the long exact sequence in cohomology you get that if two of these objects have cohomology in $$B-mod$$, so has the third (by thickness of $$D(B-mod)$$).
Conversely, $$D \subset T$$: you can proceed by induction on the length of the complex. For example, for a length two complex $$X^\bullet := 0 \to M \xrightarrow{f} N \to 0$$ in $$D$$, you have the triangle $$Y^\bullet \to X^\bullet \to Z^\bullet$$ with $$Y^\bullet = 0 \to ker(f) \to 0 \to 0$$ and $$Z^\bullet = 0 \to 0 \to coker(f) \to 0$$. This triangle is distinguished because it is isomorphic to the triangle $$Y^\bullet \to X^\bullet \to Cone$$
No the factorization $$D(B-mod) \to D \to D(A-mod)$$ is just looking at complex of $$B$$-modules as complex of $$A$$-modules with cohomology of $$B$$-modules.