[flang] Don't add a source range to a type instantiation's scope

[klausler]

Summary

A type instantiation's scope doesn't require a source range, as we don't
need/want it to be the result of SemanticsContext::FindScope(), and
adding the original type's source range to the scope of one of its
instantiations has the side effect of expanding the source range of
all of its parents to include the original type definition, which
breaks FindScope() for any scope including a type instantiation.

[clementval]

LGTM

flang/lib/Semantics/type.cpp

contextMessage = new parser::Message{foldingContext.messages().at(),
         "instantiation of parameterized derived type '%s'"_en_US, desc};
     if (auto outer{containingScope.instantiationContext()}) {
       contextMessage->SetContext(outer.get());
     }
     newScope.set_instantiationContext(contextMessage);
   }
   // Instantiate every non-parameter symbol from the original derived
   // type's scope into the new instance.
-  newScope.AddSourceRange(typeScope.sourceRange());
   auto restorer2{foldingContext.messages().SetContext(contextMessage)};
   if (PlumbPDTInstantiationDepth(&containingScope) > 100) {
     foldingContext.messages().Say(
         "Too many recursive parameterized derived type instantiations"_err_en_US);
   } else {
     InstantiateHelper{newScope}.InstantiateComponents(typeScope);
   }
 }