Rework the explanation of reference types / interfaces on the C# type system overview. (#26419)

* Rework the explanation of reference types / interfaces on the C# type system overview. 
- Add link to reference types page
- Update associated code sample to use clearer variable names.

* Missed one name change for DeclarationAndAssignment example.

* Move interface examples to own method to avoid duplicate name conflict.

* Update wording per review comments

Author: forensicmike

docs/csharp/fundamentals/types/index.md

@@ -114,11 +114,13 @@ All enums inherit from <xref:System.Enum?displayProperty=nameWithType>, which in
 
 ### Reference types
 
-A type that is defined as a `class`, `record`, [`delegate`](../../language-reference/builtin-types/reference-types.md), array, or [`interface`](../../language-reference/keywords/interface.md) is a *reference type*. At run time, when you declare a variable of a reference type, the variable contains the value [`null`](../../language-reference/keywords/null.md) until you explicitly create an object by using the [`new`](../../language-reference/operators/new-operator.md) operator, or assign it an object that has been created elsewhere by using `new`, as shown in the following example:
+A type that is defined as a `class`, `record`, [`delegate`](../../language-reference/builtin-types/reference-types.md), array, or [`interface`](../../language-reference/keywords/interface.md) is a [`reference type`](../../language-reference/keywords/reference-types.md).
+
+When declaring a variable of a [`reference type`](../../language-reference/keywords/reference-types.md), it contains the value [`null`](../../language-reference/keywords/null.md) until you assign it with an instance of that type or create one using the [`new`](../../language-reference/operators/new-operator.md) operator. Creation and assignment of a class are demonstrated in the following example:
 
 :::code language="csharp" source="../../programming-guide/types/snippets/index/Program.cs" ID="DeclarationAndAssignment":::
 
-An interface must be initialized together with a class object that implements it. If `MyClass` implements `IMyInterface`, you create an instance of `IMyInterface` as shown in the following example:
+An [`interface`](../../language-reference/keywords/interface.md) cannot be directly instantiated using the [`new`](../../language-reference/operators/new-operator.md) operator. Instead, create and assign an instance of a class that implements the interface. Consider the following example:
 
 :::code language="csharp" source="../../programming-guide/types/snippets/index/Program.cs" ID="InterfaceDeclaration":::
 

docs/csharp/programming-guide/types/snippets/index/Program.cs

@@ -132,14 +132,22 @@ static void Declarations()
         static void DeclarationAndAssignment()
         {
             // <DeclarationAndAssignment>
-            MyClass mc = new MyClass();
-            MyClass mc2 = mc;
+            MyClass myClass = new MyClass();
+            MyClass myClass2 = myClass;
             // </DeclarationAndAssignment>
+        }
 
+        static void InterfaceAssignment()
+        {
             // <InterfaceDeclaration>
-            IMyInterface iface = new MyClass();
-            // </InterfaceDeclaration>
+            MyClass myClass = new MyClass();
 
+            // Declare and assign using an existing value.
+            IMyInterface myInterface = myClass;
+
+            // Or create and assign a value in a single statement.
+            IMyInterface myInterface2 = new MyClass();
+            // </InterfaceDeclaration>
         }
 
         static void ArrayDeclaration()