Suppose I have a monad $M_S = \langle S , \eta_S, \mu_S \rangle$ and a monad map given by a natural transformation $\phi: T \rightarrow S$. What are the natural transformations of the resulting monad?
$$\langle S , \eta_S, \mu_S \rangle \rightarrow \langle T , ?, ? \rangle $$
I am looking at Street 72 and in it he defines a monad functor. I am using this paper to get my definition of a monad map by saying that the functor $U$ is identity.
Edit: someone is now stating that I have to fill in the natural transformations of the image monad, like this
$$\langle S , \eta_S, \mu_S \rangle \rightarrow \langle T , \eta_T, \mu_T \rangle $$
So, where do we start when defining a monad map? Do we say that we have two monads and IF you have a natural transformation $\phi$ as above, you have a monad map?

The following definition appears in Toposes, Triples and Theories by Barr and Wells, Section 3.6 (page 110), as well as in Dense morphisms of monads by Karazeris and Velebil, Section 2.1 (page 4).
Here the natural transformation $α α$ is given by $α S ∘ T α$, or equivalently by $S α ∘ α T$.
This definition only deals with monads on the same category. For monads on two different categories, the following can be used:
The first definition is a special case of the second definition, by considering for $F$ the identity functor.
The above definitions apply to monads on categories. However, the notion of a monad can be generalized to objects in $2$-categories by replacing categories with objects, functors with morphisms, and natural transformations with $2$-morphisms. (Instead of “objects”, “morphisms” and “$2$-morphisms” one can also use the terms “$0$-cells”, “$1$-cells” and “$2$-cells”.) The previous notion of a monad on a category is then the same as a monad in the $2$-category $\mathsf{Cat}$.
Definitions 1 and 2 can then also be applied to monads in $2$-categories. The resulting definitions of morphisms between monads in $2$-categories can be found in the nLab, as well as in The formal theory of monads by Ross Street, which was already mentioned in the question.