The answer is - it does! Be careful, your formula for E only works because of spherical symmetry.
The electric field at a point in space between the two spheres due to a negative charge on the outer sphere is perfectly cancelled due to the combined contributions of negative charge distributed over the rest of the outer sphere.
The second field calculation uses Gauss's law, which does not work as you have stated, instead it says that the flux of electric field out of a closed surface equals the charge (divided by $\epsilon_0$ contained within. Only in a case of high symmetry will you be able to say that the electric field cuts the spherical surface perpendicularly and that the flux is $4\pi r^2 E$.
If however, the two shells you have drawn are conductors it will always be the case that the interior (static) E-field due to charges on the outer conductor will be zero. i.e. To answer your question - you will always be able to neglect the interior field from the outer conductor because it is zero. This will apply for any shape outer conductor. The E-field at the surface of any conductor must be parallel to the surface vector or free charges would move to make it so. In that case you can always construct a Gaussian surface with the E-field parallel to the surface vector and, because there is no enclosed charge, the next flux must be zero, so the E-field must be zero. In other words, the (static) E-field is always zero inside a hollow conductor. So the only field that need be taken account of is that produced by the inner conductor (recall that you can simply superpose electric fields).
(Aside: I specify static, because it is possible for time-variable E-fields to penetrate inside a conducting shell if it is thin enough, by inducing time-varying currents in the shell).
(A) When using Gauss' law for spherically symmetric systems, $r$ is evaluated at the radius of the Gaussian surface. I understand where the confusion comes from, because in the "standard" equation for the electric field, the distance $r$ is defined as the distance between the point where you're measuring the field and the point where the charge is located, and in the case of a spherical surface, you might think that measuring the field on the surface that since you are infinitesimally close to the surface, $r$ should be zero. In a real system, if you were able to get that close to the surface, you would actually have to start worrying about where the quantized charges are located, so things would get complicated close to the surface. For these types of problems, it is assumed that the charges are "smoothed" out over the surface, and using Gauss' law comes in handy. Gauss' Law is used in the following way.
$\oint \vec{E}\cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$
Since you chose a spherical surface $\vec{E}$ and $d\vec{a}$ are parallel. Also, $\vec{E}$ is constant over the surface, so the integral is just the following:
$E\oint da = \frac{Q_{enc}}{\epsilon_0}$
where $\oint da= 4\pi r^2$, and $r$ is the radius of the Gaussian surface. So,
$\vec{E} = \frac{Q_{enc}}{4\pi r^2} \hat{r}$
Now, you can see that $r$ is the radius of the sphere, and the only place that the size of the charged sphere would matter is when determining whether the charge on the surface of the sphere is enclosed or not. Since on a conductor, all of the charge is on the outer surface, at the surface itself, the charge is generally not considered to be enclosed, but just above the surface (infinitesimally), the charge is considered to be enclosed (all of the charge on the sphere).
(B) In this case, you have to remember that conductors do not like to have electric fields on the inside. So when connecting the two sphere's by a conduction wire, any net charge will want to move to the outer surface. So you are right in this case, the charge on the inner sphere is zero and the net charge is on the outer sphere.
(C) The work done by an electric field on a particle is $q\Delta V$. The way to think of it is that the work done on something is equal to the change in potential energy of that object. In this case, the change in the potential energy of a charge is given by $\Delta U = q\Delta V$. Now, the sign of $q$ and $\Delta V$ should be determined by the system.
For instance lets consider free charges at rest in an electric field. A proton has $q=+e$ and an electron has $q=-e$. Protons move from high potential to low potential, and electrons move from low potential to high potential. So, with $\Delta V = V_f - V_i$, in both cases you see a negative work done by an electric field, because the charges will move toward a position that lowers their potential energy. In order to do (positive) work on a charge, you need to raise the potential energy. This is something that batteries do.
Best Answer
Quite a lot of questions, but I think they can all be answered by understanding the main argument: 'electric field lines can not pass through a conductor'. I will illustrate with the spherical shell-question.
The statement that electric field lines cannot pass through a conductor, is simply wrong. The reason that your flashlight works, is that the battery produces an electric field in the wiring, which causes free electrons to move and convert their kinetic energy into light and heat in the bulb.
However, when there is a static equilibrium (no moving charges), it is true that the electric field inside a conductor must be zero. If the electric field wouldn't be zero inside a conductor, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. This doesn't apply to insulators, since there, the charged particles can't move. (The same reasoning holds for 'why is the angle between electric field and a conductor always 90 degrees').
Now, the rearranging means that (in the case of a positive charge inside), the free electrons will go to the inner side of the shell, so that within the shell, the total electric field is zero: $$\vec{E} = \vec{E}_{q,inside}+\vec{E}_{free \ electrons}=\vec{0}$$ This of course leads to an induced charge on the outer side of the shell, which creates an electric field outside the shell:
$$\vec{E} = \vec{E}_{q,inside}+\vec{E}_{free \ electrons}+\vec{E}_{induced}=\vec{E}_{induced}$$
But, as the induced charge is equal to q, you can act as if the shell isn't there when calculating the field outside the shell (just calculate $E_q$).