By Symmetry's answer is correct, but I don't think it answers quite the question that you're getting at.
You are making a very common conceptual mistake for students first learning about quantum statistics, which is that quantum particles "need" to be indistinguishable and that (anti)symmetrization is the way that they "satisfy that need." But this isn't the case - one could certainly imagine a quantum system of multiple particles with completely identical mass, charge, spin, etc., but where the particles are "distinguishable" in the sense that their joint wavefunction is neither symmetric nor antisymmetric under particle exchange. Such particles would be neither bosons nor fermions. (In fact, condensed matter physicists do this all the time when they consider magnetic spin systems, and less commonly, systems of "anyons" which are neither boson nor fermionic.)
So why do intro QM textbooks (almost) always only consider bosonic or fermionic particles? Because every single fundamental particle in the Standard Model has been experimentally shown to be either a boson or a fermion. Recall that any physical system has an associated "Hilbert space," which is the set of quantum states in which it is physically possible to find the system. The set of totally symmetric wavefunctions forms the "bosonic Hilbert space" and the set of totally antisymmetric wavefunctions forms the "fermionic Hilbert space," and all known species of elementary particles are described by one or the other space. But a more general Hilbert space would certainly be logically and mathematically consistent.
(When you study QM at a more advanced level, you'll learn that there are deeper reasons why bosonic and fermionic theories are more "natural." In these two special cases, we can use a very elegant mathematical formalism called "second quantization" that saves us a lot of work. Spin and anyonic systems are in some ways more difficult to deal with than systems of bosons and fermions, because you can't use second quantization - or sometimes you can use it, but in a much, much more complicated way, by mapping the system to a bosonic or fermionic "gauge theory." Also, there is no known way to make a particle that is neither bosonic nor fermionic compatible with special relativity.)
Anyway, this is a long-winded way of saying that all multi-electron wavefunctions are always antisymmetrized, no matter how far away they are. (Although it turns out that you can't use this antisymmetrization to send information faster than light, so special relativity is safe.) But recall that in quantum mechanics, only inner products are physically observable, not the wavefunctions themselves. (Actually, only the norm-squared of an inner products is physically observable.) If the particles don't have any spatial overlap, then it turns out that if we want to evaluate $\langle \hat{X} \rangle$, we'll get the same answer whether or not we use the correct, antisymmetrized wavefunction, or a hypothetical unsymmetrized wavefunction. So even though the unsymmetrized wavefunction doesn't even lie in the Hilbert space and doesn't make any sense physically, we can get away with using it even though it's "wrong." Try it! Write down two non-overlapping wavefunctions and calculate $\langle \hat{X} \rangle$ with respect to both the antisymmetrized and non-antisymmetrized states - you'll get the same answer either way. So really, strictly speaking you still "have to" antisymmetrize, but you can get away with "cheating" and neglecting to do so if the particle are far away and have negligible spatial overlap.
Finally, you might wonder "if all known fundamental particles are either bosons or fermions, then why do condensed matter physicists ever bother to consider these weird magnetic spin and anyonic systems?" The answer is actually different in the two cases. Magnetic spin systems are actually electron systems where every electron's wavefuncton falls off so fast that we can treat them as "far away," even if they're only separated by an interatomic spacing! So we can get away with ignoring the antisymmetrization, even though it's "really" there. Anyons are even stranger and don't correspond to any individual fundamental particles at all - they are "collective excitations" that only emerge when you take ginormous numbers of electrons and couple them together in very special ways.
The requirement is
$$
\psi(x_1,x_2) =
\begin{cases}
\psi(x_2,x_1) & \text{for bosons} \\
-\psi(x_2,x_1) & \text{for fermions}.
\end{cases}
\tag{1}
$$
This property is required by the spin-statistics theorem in relativistic quantum field theory. Since non-relativistic quantum mechanics is supposed to be an approximation to relativistic quantum field theory, we also enforce it in non-relativistic QM.
A special case of equation (1) is
$$
\psi(x_1,x_2) \approx
\begin{cases}
f(x_1)g(x_2)+f(x_2)g(x_1) & \text{for bosons} \\
f(x_1)g(x_2)-f(x_2)g(x_1) & \text{for fermions},
\end{cases}
\tag{2}
$$
but like Lewis Miller's answer said, this is only a special case. The general requirement is equation (1).
The square-root example written in the question does not satisfy the requirement (1).
Best Answer
The probability density to find one particle at $r_1$ and the other particle at $r_2$ is the absolute square of the wave-function. The question is, so I understand, how does the wave-function look like. If the particles a and b are distinguishable, the wave-function looks as given in the exercise. However, if the particles are indistinguishable fermions, the wave function should be antisymmetrical at the interchange of the particles, i.e. you do antisymmetrization:
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) - ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$
On the other hand, if the particles are indistinguishable bosons, you do symmetrization
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) + ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$