It does not; and if you call super() more than once it will actually throw an error.
You got it the wrong way round here: you have to assign this.varA = varA so that’s available to subclasses and other methods. And any subclass will have to actually pass that argument to the parent constructor, e.g. by just forwarding its own constructor arguments like
class A {
constructor (varA) {
this.varA = varA
}
}
class B extends A {
constructor (...args) {
super(...args)
console.log(this.varA)
}
}
const b = new B('foo') // logs "foo"