It helps to take a look in scala.Predef to see what exactly is going on.
If you check there, you see that String in Scala is just an alias for java.lang.String. In other words, the + method on a String gets translated into Java's + operator.
So, if a Scala String is just a Java String, how does the ++ method even exist, you might ask. (Well, I'd ask, at least.) The answer is that there's an implicit conversion from String to WrappedString provided by the wrapString method, which is also in Predef.
Notice that ++ takes any GenTraversableOnce instance and adds all of the elements in that instance to the original WrappedString. (Note that the docs incorrectly state that the method returns a WrappedString[B]. This has to be incorrect, because WrappedString doesn't take type parameters.) What you'll get back is either a String (if the thing you add is a Seq[Char]) or some IndexedSeq[Any] (if it's not).
Here are some examples:
If you add a String to a List[Char], you get a String.
scala> "a" ++ List('b', 'c', 'd')
res0: String = abcd
If you add a String to a List[String], you get an IndexedSeq[Any]. In fact, the first two elements are Chars, but the last three are Strings, as the follow-up call shows.
scala> "ab" ++ List("c", "d", "e")
res0: scala.collection.immutable.IndexedSeq[Any] = Vector(a, b, c, d, e)
scala> res0 map ((x: Any) => x.getClass.getSimpleName)
res1: scala.collection.immutable.IndexedSeq[String] = Vector(Character, Character, String, String, String)
Finally, if you add a String to a String with ++, you get back a String. The reason for this is that WrappedString inherits from IndexedSeq[Char], so this is a convoluted way of adding a Seq[Char] to a Seq[Char], which gives you back a Seq[Char].
scala> "abc" + "def"
res0: String = abcdef
As Alexey noted, neither of these is a very subtle tool, so you're probably better off using string interpolation or a StringBuilder unless there's a good reason not to.