$\let\d=\delta \let\dag=\dagger \def\ket#1{|#1\rangle}
\def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle}
\def\bp{{\mathbf p}} \def\br{{\mathbf r}}$
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 $2p^0$.
