Electric field on Gaussian surface due to external charge 
Gauss' law gives me the electric field at point P due to the charge +q enclosed by the Gaussian surface, but what if there is an external charge +Q? Won't it influence the electric field at point P, as electric field vector at that point due to both the charges will get added according to vector laws? This confuses me, does the Gauss law give me the electric field at point P due to the charge the Gaussian surface encloses or does it give me the net electric field , that is the resultant of electric fields of both +q and +Q?
 A: Gauss's law in general relates the total electric field flux through a closed surface to the charge enclosed in that surface. It does not give you the field at a certain point in space in general. You can, however, exploit symmetries in certain geometries (points, spheres, lines, cylinders, planes, etc.) to determine the field at certain points in space. If you can't exploit some symmetry to pull E out of your flux integral, then you can't use Gauss's law to determine the field at a point in space.
It's like saying "I have 10 numbers that add to 100." If this is all you know then you can't tell me which numbers I have used to get a total of 100. But if I also tell you "each number is the same number", then you can for sure say I used 10 10's to get a total of 100.
As for your diagram, yes the charge Q will influence the field at P. It will not change the total flux through your Gaussian sphere, or any other Gaussian surface not enclosing Q. You will not be able to use Gauss's law now to find the field at P since the symmetry that lets you pull E out of the integral is no longer present. (Although, you could argue that you can use Gauss's law on each charge individually and then use superposition to get the total field, but you can't use Gauss's law for the entire configuration to get the field at P in this case).
