Force from a wavefunction/superposition on other particles How does force-interaction which is dependent on some uncertain property(like position or velocity) work in QM? If I have a charged particle described by some position wavefunction, how would I expect the EM force on other particles/wavefuctions to work? Does the force come from a sort of "smeared out" area of space?
 A: "Does the force come from a sort of "smeared out" area of space?"  That is the approximation usually employed (called mean field approximation).  In QM we normally don't deal with forces, instead we study potential functions.  Of course classically forces are defined as the gradient of potential functions, but as you indicate the uncertainty associated with QM makes this concept somewhat vague.  There is a procedure for many-electron systems (called the Hartree-Fock approximation) wherein each electron feels the EM force from the nuclei present as well as the "smeared out" EM potential from all the other electrons (calculated from their probability distributions).  This results in a numerical problem for each electron (or orbital) and the result is iterated until the electron distributions in two successive iterations are identical to some level of precision.  At this point the results are called self-consistent (furthur iterations would yield identical wave functions).
A: Particles are in the realm of quantum mechanics, and in quantum mechanics one does not have "forces" on par with classical mechanics forces, one has a dp/dt momentum transfere in an interaction of two particles, and that is the force.
For a simple example two electrons interacting are depicted by Feynman diagrams, which are a pictorial representation for the integrals which will predict the behavior of the scattering:

At the vertex a dp/dt is exchanged between the two incoming electrons, and that is the force in quantum mechanics, which changes the directions of the two outgoing electrons and the distribution of the scattering angle ( do not forget we are talking probabilities at the quantum level) is calculated to first order by the recipe that turns feynman diagrams into calculations. It is very deterministic, but what is determined is the probability distribution for the scatter.
A: The idea of a "force" acting between two particles is more useful when both particles are sharply localized. It is less useful when they are not sharply localized. A more general concept is how the two particles influence each other, or — even better — how their mutual interaction affects the behavior of the overall two-particle wavefunction. From this more-general perspective, we can recover the usual idea of a "force" under the special conditions where that idea is useful.
To illustrate this, suppose we have two charged particles, say an electron and a proton. To keep things as simple as possible, consider a nonrelativistic model in which these particles interact with each other only via the Coulomb interaction, with no quantum EM field, and ignore the particles' spin. Then the state of the system is described by a wavefunction $\psi(x_e,x_p)$, where $x_e$ corresponds to the electron and $x_p$ to the proton.
If the electron and proton are both sharply localized (with narrow wavefunctions) compared to the distance between them, and if their momenta are also both sharply localized (as much as allowed by the uncertainty principle), then the mutual interaction between the electron and proton will cause them to accelerate toward each other much like two classical particles. This is an approximation that breaks down when the distance between them is no longer large compared to the widths of their wavefunctions.
To get some intuition into what happens when this approximation breaks down, consider the situation depicted below:

The top part of the picture depicts an initial state of the form
$$
   \psi(x_e,x_p)=A(x_e)\big(B_L(x_p)+B_R(x_p)\big) = 
   A(x_e)B_L(x_p)+A(x_e)B_R(x_p),
\tag{1}
$$
where $A$ is the electron wavefunction (the blue packet in the middle of each rectangle) and $B_L$ and $B_R$ are two different wavepackets for the proton, one localized to the left of the electron (as shown in the left rectangle) and one localized to the right of the electron (as shown in the right rectangle). In words, the proton's initial wavefunction has two localization regions, one on either side of the electron. The time-evolution of the wavefunction (1) is described by the two-particle Schrödinger equation, which is linear, which means that the result after some time has passed will be the same as time-evolving each of the two terms individually and them adding them together again in the end:


*

*In the first term of the superposition (the left half in the picture), the proton pulls the electron to the left.

*In the second term of the superposition (the right half in the picture), the proton pulls the electron to the right.
In the resulting final state, the electron's location has become entangled with the proton's location: the final state is a quantum superposition of "proton and electron are both off to the left" and "proton and electron are both off to the right":
$$
   \psi(x_e,x_p)=
   A_L(x_e)B_L(x_p)+A_R(x_e)B_R(x_p),
\tag{2}
$$
where $A_L$ and $A_R$ are two differently-localized electron wavepackets. This is still a two-particle state (one electron and one proton), but neither particle has a location of its own, not even approximately. The two-particle wavefunction can no longer be written as a product of two single-particle wavefunctions; their locations have become entangled with each other. 
This illustrates how the electron is affected when the proton is delocalized. This example was contrived to be simple. More generally, the proton won't be in a superposition of two sharply-defined locations like this; it will be spatially spread out in some more general way, and then translating the math into useful intuition becomes more difficult.
Even in the original scenario where the proton was sharply localized near a single location, the electron's and proton's wavefunctions still have some finite width, so they will still become entangled with each other to some degree. This becomes more pronounced as they get closer to each other, when the distance between them is no longer much larger than the widths of the wavepackets. This happens partly because the Coulomb interaction is pulling them together, and partly because their wavefunctions (even if there were no interaction at all) tend to spread out as time passes — because their momenta are not perfectly sharply defined, either. 
