Except the 'a' in 'A::B::a' (if used as a name), or in your first example's 'T::U::a', is not part of any nested-name-specifier. See the syntax in [expr.prim.id.qual]:

qualified-id:

    nested-name-specifier template_opt unqualified-id

nested-name-specifier:

    ::

    type-name ::

    namespace-name ::

    decltype-specifier ::

    nested-name-specifier :: identifier

    nested-name-specifier :: template_opt simple-template-id ::

(Later drafts replace decltype-specifier with computed-type-specifier to also deal better with variadic parameter packs.)

The wording about a nested-name-specifier containing a nested-name-specifier and its identifier or simple-template-id refers to rules 4 and 5 for the grammar node nested-name-specifier.

In your first example, 'T::U::a' is a qualified-id containing the nested-name-specifier 'T::U::' and the unqualified-id 'a'. And that nested-name-specifier 'T::U::' uses the fourth grammar rule, immediately containing another type-dependent nested-name-specifier 'T::' and the identifier 'U'. So breaking down the quoted requirement,

"In a nested-name-specifier"  (T::U::)

"that immediately contains a nested-name-specifier"  (T::)

"that depends on a template parameter,"  (the type T in T:: is type-dependent)

"the identifier"  (U)

"or simple-template-id"   (If we had 'T::template X<int>::b' using grammar rule 5, 'X<int>' would be the simple-template-id.)

"is implicitly assumed to name a type"  (T::U:: names a type, to be determined at template instantiation.)
"without the use of the typename keyword"  (which is good, since 'T::typename U::a' is invalid syntax and 'typename T::U::a' makes 'a' a type, not 'U')

-- Andrew Schepler