Flask-Sqlalchemy multiple databases (binds) with the same model (class DB.Model)

I'm a beginner with python/Flask/SQLAlchemy so sorry if my questions are dumb.

I want to create an API with Flask using Flask-SQLAlchemy as following:

1. one sqlite database for users/passwords

SQLALCHEMY_DATABASE_URI = 'sqlite:////path/to/users.db'

 class User(DB.Model):
    __tablename__ = 'users'
    id = DB.Column(DB.Integer, primary_key=True)
    username = DB.Column(DB.String(64), index=True)
    password = DB.Column(DB.String(128))

2. Lets say I have multiple "customers" witch a user can create using

   $ http POST http://localhost:5000/api/customers/ name=customer1

   I would have a model like this:

   class CustomerModel(DB.Model):

   __tablename__ = 'customer'
   customer_id = DB.Column(DB.Integer, primary_key=True)
   customer_name = DB.Column(DB.String, unique=True, index=True)
   

I need to create a separate sqlite file (example: customer1.db) for each "customers" so I would have:

SQLALCHEMY_BINDS = {
    'customer1' = 'sqlite:////path/customer1.db',
    'customer2' = 'sqlite:////path/customer2.db',
    ...
}

My questions are:

1. I do not have fixed number of "customers" so I cannot create a model class for each and specify the "bind_key" for each. Is it possible to do this with Flask-SQLAlchemy or I need to use plain SQLAlchemy?

2. lets say I have 3 "customers" in data/ as customer1.db, customer2.db and customer3.db. I would start the application and create SQLALCHEMY_BINDS dictionary listing the files in data/ and then DB.create_all().

on a request for a specific "customer" how can I bind to the right .db file using the Flask-SQLAlchemy DB.session? I've seen this but there is no answer and I cannot find the way to do it. Using different binds in the same class in Flask-SQLAlchemy