I haven't checked all the details, but I think the story could go like this. (I have to apologize: it's a bit long.)

(1) Let F: **A** ---> **B** be an additive left exact functor between two abelian categories. Take an injective resolution of an object A in **A** :

0---> A ---> I^0 ---> I^1 ---> ...

Let us call i: A ---> I^0 the first morphism. Apply F to this exact sequence:

0---> FA ---> FI^0 ---> FI^1 ---> ...

Now, the *total* right derived functor of F applied to A (thought as a complex concentrated in degree zero) is the complex

**R**F(A) = { FI^0 ---> FI^1 ---> FI^2 ---> ... }

and the *classical* right derived functors of F are its cohomology:

R^nF(A) = H^n(**R**F(A)) = H^n(FI^*) .

These {R^nF}_n are a universal cohomological delta-functor and we have a natural transformation of functors

qF ===> (**R**F)q

which is essentially

Fi: FA ---> **R**F(A)

(here we have extended F degree-wise to the category of complexes, and this is the degree zero of the natural transformation, because **R**F(A)^0 = FI^0 ).

(2) Now, let { T^n : **A** ---> **B** } be a cohomological delta-functor and f^0 : F ---> T^0 a natural transformation. We have to extend this f^0 to a unique morphism of delta-functors { f^n : R^nF ---> T^n } .

To do this, observe that, in general, given two right-derivable functors between two, say, model categories

F, G: **C** ---> **D** ,

and a natural transformation between them

t: F ===> G

we have a natural transformation between the total right derived functors

**R**t : **R**F ===> **R**G

because of the universal property of the derived functors: indeed, if

f : qF ===> (**R**F)q and g : qG ===> (**R**G)q

are the universal morphisms of the derived functors, then we have a natural transformation

gt : F ===> (**R**G)q

and, so, because of the universal property of derived functors, a unique natural transformation

**R**t : **R**F ===> **R**G

such that (**R**t)qf = g .

(3) So, take our f^0 : F ---> T^0 , extend it to a natural transformation between the degree-wise induced functors between complexes. Passing to the derived functors, we obtain

**R**f^0 : **R**F ===> **R**T^0 .

Taking cohomology, for each n , we get

H^n(**R**f^0) : H^n (**R**F) ===> H^n (**R**T^0) .

But these are the classical right derived functors, so we have natural transformations

R^nf : R^n F ===> R^nT^0

and because the classical right derived functors are universal delta-functors, we have unique natural transformations

i^n : R^nT^0 ===> T^n

which extend the identity

i^0 : R^0T^0 = T^0 .

The composition

i^n R^f : R^F ===> T^n

is, I think, the required morphisms of delta-functors that we need.