Shapely Distance different from Geopy/Haversine

Because the principles and the algorithms are different (look at Geographical distance)

1. Shapely use the euclidean distance in a cartesian plane and the shortest distance between two points in a plane is a straight line which contains the two points.

 import numpy as np
 print np.linalg.norm(np.array(pt_user) - np.array(pt_store))
 110.02637304449682 # meters
 from scipy.spatial import distance
 print distance.euclidean(pt_user, pt_store)
 110.02637304449682 # meters

2. Vincenty, Great Circle and Haversine use either the geodesic distance (on an ellipsoid, Vincenty) or the great-circle distance (the shortest distance along the surface of a sphere) between two points. The shortest distance on the surface of a sphere is along the great-circle which contains the two points.

   Therefore it is normal that the Shapely, Numpy and Scipy euclidean distances differ from the Vincenty, Great Circle and Haversine distances and the differences between the Vincenty, Great Circles and Haversine distances are linked to the choice of an ellipsoid, and many other things.

   You can also change the ellipsoid

 print vincenty((39.435307,-76.799614),(39.43604,-76.79989),ellipsoid='WGS-84')
 0.0847784769149 km
 print vincenty((39.435307,-76.799614),(39.43604,-76.79989),ellipsoid='GRS-80')
 0.0847784769128 km

Or use other libraries as geodistance

 print geodistance.distanceVincenty(39.435307,-76.799614,39.43604,-76.79989, ellipsoid='WGS-84')
 (0.08477847691523362, -16.276730447136675) # distance, azimuth
 print geodistance.distanceHaversine(39.435307,-76.799614,39.43604,-76.79989)
 (0.08488248586585143, -16.214988211007256)

You can see that all the differences are centimetric. With metric precision, all the values = 85 meters.

3. Which one is right? All, because it depends on the context: if you work with projected data (cartesian plane), you use the Euclidean distance (Shapely, Numpy ,Scipy and many others), if not, one of the others.

   They are also many other distances (Scipy Spatial distances)

New

In support of the answer of Mintx

pt_store=Point(transform(Proj(init='EPSG:4326'),Proj(init='EPSG:31370'),-76.799614, 39.435307))
pt_user=Point(transform(Proj(init='EPSG:4326'),Proj(init='EPSG:31370'),-76.79989,39.43604))
pt_store.distance(pt_user)
86.26511001003892

And here's another distance calculation from GeographicLib:

from geographiclib.geodesic import Geodesic
g = Geodesic.WGS84.Inverse(39.435307, -76.799614, 39.43604, -76.79989)
print(g['s12'])  # 84.7784769689

I would consider this the right one, within 15 nanometres.