How to use the request in a ModelForm in Django

No, the request is not passed to the ModelForm. You'll need to do something like this in your view:

form = BookSubmitForm()
form.fields['book'].queryset = Book.objects.filter(owner=request.user)
# pass form to template, etc

As you said, it's often cleaner to encapsulate this in the Form object, particularly if you have several fields that will need filtered querysets. To do this, override the forms's __init__() and have it accept a kwarg of request:

class BookSubmitForm(ModelForm):
    def __init__(self, *args, **kwargs):
        self.request = kwargs.pop("request")
        super(BookSubmitForm, self).__init__(*args, **kwargs)
        self.fields["book"].queryset = Book.objects.filter(owner=self.request.user)
        self.fields["whatever"].queryset = WhateverModel.objects.filter(user=self.request.user)

Then just pass request whenever you instantiate BookSubmitForm in your view:

def book_submit(request):
    if request.method == "POST":
        form = BookSubmitForm(request.POST, request=request)
        # do whatever
    else:
        form = BookSubmitForm(request=request)
    # render form, etc

Extending AdamKG answer to class based views - override the get_form_kwargs method:

class PassRequestToFormViewMixin:
    def get_form_kwargs(self):
        kwargs = super().get_form_kwargs()
        kwargs['request'] = self.request
        return kwargs

from django.views.generic.edit import CreateView
class BookSubmitCreateView(PassRequestToFormViewMixin, CreateView):
    form_class = BookSubmitForm
# same for EditView

and then in forms:

from django.forms import ModelForm
class BookSubmitForm(ModelForm):
    def __init__(self, *args, **kwargs):
        self.request = kwargs.pop("request")
        super().__init__(*args, **kwargs)
        ...