Transition to new `split_line` implementation

[ghendrickx]

Within the repository, there were multiple functions that split a line (shapely.LineString) by a point (shapely.Point). In #494, a robust line-splitting function has been implemented (64f02db-846ef2d), replacing other/older implementations. The new function gets in robustness by not relying on the shapely.ops.split-function for splitting the line by a point, which is known not to work robustly using this built-in split-function.

Due to the modifications - and aligning the various line-splitting function - the signature slightly changes and the intended output will change as well:

1. Signature change from def split_line(line: shapely.LineString, point: shapely.Point, tolerance: float = 0.1) (or similar) to def split_line(line: shapely.LineString, point: shapely.Point, *, tolerance: float = 0.1) (excluding transitional arguments).
2. Output change from shapely.MultiLineString | shapely.LineString to tuple[shapely.LineString, ...].

The code-changes in #494 already include a transition strategy by a temporary signature (and typing.overload-functions):

@typing.overload
def split_line(
    line: LineString, point: Point, *deprecated: typing.Any, tolerance: float = 0.1, as_multilinestring: None
) -> MultiLineString | LineString: ...


# TODO: Next transition step: Default `as_multilinestring=None` with `=True`-behaviour and a UserWarning


@typing.overload
def split_line(
    line: LineString,
    point: Point,
    *deprecated: typing.Any,
    tolerance: float = 0.1,
    as_multilinestring: typing.Literal[True] = True,
) -> MultiLineString | LineString: ...


@typing.overload
def split_line(
    line: LineString,
    point: Point,
    *deprecated: typing.Any,
    tolerance: float = 0.1,
    as_multilinestring: typing.Literal[False],
) -> tuple[LineString, ...]: ...


def split_line(
    line: LineString,
    point: Point,
    *deprecated: typing.Any,
    tolerance: float = 0.1,
    as_multilinestring: bool | None = True,
) -> MultiLineString | LineString | tuple[LineString, ...]:
    ...

The *deprecated-argument catches positional use of the tolerance-argument, which will become keyword-only:

    if deprecated:
        warnings.warn(
            "Passing 'tolerance' positionally is deprecated. Use keyword argument 'tolerance='.",
            FutureWarning,
            stacklevel=2,
        )
        tolerance = deprecated[0]

The as_multilinestring-argument defaults to the old behaviour of returning shapely.MultiLineString-objects (or shapely.LineString-objects). Currently, it defaults to True. The next transition step would have it change to None by default; next, defaulting to False; and eventually will be removed and the function will always return the intended, more robust output: tuple[shapely.LineString, ...]:

    if as_multilinestring is None:
        warnings.warn(
            "Calling 'split_line' without specifying 'as_multilinestring' is deprecated. "
            "In a future version, this will return 'tuple[LineString, ...]'. "
            "Set 'as_multilinestring' explicitly (and set to False for future-proofing the code).",
            UserWarning,
            stacklevel=2,
        )
    elif as_multilinestring:
        warnings.warn(
            "The 'split_line'-function will return tuple with LineString-objects in the future. "
            "Change to the future behaviour by setting 'as_multilinestring=False'.",
            FutureWarning,
            stacklevel=2,
        )