How should I imagine a multi-particle state in a free QFT? It is reasonable to think of single-particle Focks states as of plane waves. Indeed, since $|p\rangle=a^\dagger_p|0\rangle$ and $\langle x|p\rangle\sim \operatorname{e}^{ipx}$, we conclude that the state $|p\rangle$ can be thought of as a plane wave in the position representation.
What about multi-particle states, such as $|p_1,p_2\rangle=a^\dagger_{p_1}a^\dagger_{p_2}|0\rangle$? Naturally, I would imagine those to be superpositions of plane waves (since in the free theory we are dealing with equations obeying the superposition principle). But is it really like that? Is the matrix element $\langle x|p_1,p_2\rangle$ really equal to the sum of plane waves?
 A: $\let\d=\delta \let\dag=\dagger \def\ket#1{|#1\rangle} 
\def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle} 
\def\bp{{\mathbf p}} \def\br{{\mathbf r}}$

Indeed, since $\ket p = a^\dag p\ket0$ and 
  $\braket x p \sim e^{ipx}$

First. You're mixing Fock kets and usual QM kets. What is $\bra x\,$?
Second. Even though it's usual and convenient, it isn't mandatory to
assume in Fock space as one-particle base vectors the eigenvectors of
momentum. 
Before proceeding notation has to be clearly defined. You didn't say
what you mean by $p$. I'll write $\bp,$ $\br$ to mean 3-vectors,
whereas $p$ is reserved for a 4-vector.
Third. $\ket{\bp_1,\bp_2}$ is no superposition of plane waves. The
latter would be a one-particle state with momentum not well defined.
It should be written
$$\sum_\bp c(\bp) \ket \bp$$
or also as an integral, it you like. A two-particle state is
$$\ket{\bp_1,\bp_2} = a^\dag_{\bp_1} a^\dag_{\bp_2} \ket0.$$
You may also think of a general two-particle state, to be written
$$\ket{2,\xi} = \int d\bp_1 d\bp_2\,\phi(\bp_1,\bp_2)
  \ket{\bp_1,\bp_2}.\tag1$$
If your particles are bosons $a^\dag_{p_1}$ and $a^\dag_{p_2}$ commute
and the state is automatically symmetric. In the first member I wrote
a ket with a label "2" reminding it's a two-particle state, and
an arbitrary label $\xi$ to identify that state, in case it should be
necessary to distinguish from some other one.
Now let's define position eigenvector:
$$\ket\br = \int\!\!d\bp\,e^{-i\bp\cdot\br} \ket\bp \tag2$$
(normalization apart). Note that
$$\braket \br \bp = e^{i\bp\cdot\br}.$$
You can also define $a^\dag_\br$, the creation operator of a particle
in position $\br$:
$$a^\dag_\br = \int\!\!d\bp\,e^{-i\bp\cdot\br} a^\dag_\bp.$$
It's easy to verify that
$$a^\dag_\br \ket0 = \ket\br.$$
And finally, it's immediate how to write a two-particle state in
position representation. Inverting (2), always neglecting
normalization:
$$\ket\bp = \int\!\!d\br\,e^{i\bp\cdot\br} \ket\br.$$
Substituting into (1):
$$\ket{2,\xi} = \int\!\!d\br_1 d\br_2\,\psi(\br_1,\br_2)\,
  \ket{\br_1,\br_2}$$
where
$$\ket{\br_1,\br_2} = a^\dag_{\br_1} a^\dag_{\br_2} \ket0$$
and $\psi(\br_1,\br_2)$ is the Fourier transform of $\phi(\bp_1,\bp_2).$

A bit of clarification is in order as to the mass shell. @user1620696
wrote

the (free theory) states are square integrable functions on the
  positive mass shell

and my definitions above could seem in contradiction with that
statement. But let we see.
A function on the positive mass shell is a function of 4-vector $p$,
whose support is restricted to points satisfying
$$p^2 = p^\mu p_\mu = m^2,\qquad p^0 > 0.\tag3$$
Integration on positive mass shell may be written
$$\int\!\!d^{(4)}p\,F(p)\,\d(p^2 - m^2) \tag4$$
(this is not exact, since negative mass shell is not excluded, but 
is of no consequence because of the following).
From (3) we have
$$\d(p^2 - m^2) = 
  {1 \over 2 p^0}\,\d\!\left(p^0 - \sqrt{\bp^2 + m^2}\right)$$
(not exact, but negative mass shell is to be excluded)
and integral (4) becomes
$$\int\!{d\bp \over 2 p^0}\,F(\bp)$$
with
$$F(\bp) = 2\,p^0 F(p).$$
So we see that integrating on 3-space or integrating on mass shell only differs by a factor $2p^0$.
A: Previous answers look right, but I hope I may be able to add clarity.
In a free-field theory, you can read $a^\dagger_{p_1}a^\dagger_{p_2}|0\rangle$ as 'the occupation number of mode $p_1$ is 1 and the occupation number of mode $p_2$ is 1'. If you want to go to the language of spinor-valued (i.e. not operator-valued) wavefunctions in space, then the form of the wavefunction for such a state is $\frac{1}{\sqrt{2}}(\phi_A({\bf r}_1, \chi_1) \otimes \phi_B({\bf r}_2,\chi_2) + \phi_A({\bf r}_2, \chi_2) \otimes \phi_B({\bf r}_1, \chi_1))$ where A and B are the two modes, 1 and 2 are labels attached to the particles, and I assumed they are Bosons (so I wrote a symmetrised state). $\bf r$ refers to spatial position, $\chi$ refers to spin state. Now, depending on what calculations you are doing, this second notation may or may not be useful. It is basically an attempt to handle a field by using the methods of single-particle quantum theory, which can be useful sometimes, and maybe gives some physical intuition, but ultimately will not allow you to complete the kinds of calculations that particle physics needs for scattering problems etc. However it can sometimes be useful for looking at photons propagating through an optical system, or things like that.
A: Take a compound system of many strongly interacting particles at $T=0$ and give it a shake. The system in an excited state can be approximately described as a set of free quasi-particles corresponding to the normal modes of the system.
A: 
Is the matrix element ⟨x|p 1 ,p 2 ⟩ 
  ⟨x|p1,p2⟩
   really equal to the sum of plane waves?

Nope, it's the product of plane waves, not sum. 
Of cause, you have to symmetrize or anti-symmetrize the product according the statistics (boson/fermion) of the particle. In the end, it's a symmetrical/anti-symmetrical sum of products. 
