I'm using D. J. Griffiths's textbook Introduction to Quantum Mechanics (3rd ed.) for my introductory university course on the subject. In chapter 5 (starting at section 5.1.1), he discusses the behaviour of identical particles.
For a start, he introduces an elementary spatial wave function for a system of two non-interacting particles where one of them is in state $\psi_a$ and the other is in state $\psi_b$:
$$\psi(\mathbf{r_1},\mathbf{r_2})=\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})$$
He shortly after introduces how identical particles cannot be told apart, so, because "one of them" and "the other" is physically ambiguous, we write the spatial wave function of such a two-particle system as a superposition:
$$\psi_{\pm}(\mathbf{r_1},\mathbf{r_2})=A\,(\psi(\mathbf{r_1},\mathbf{r_2})\pm\psi(\mathbf{r_2},\mathbf{r_1}))$$
He posits that $\psi_+$ is the governing wave function for bosons, and $\psi_-$ for fermions $-$ which make for, respectively, a symmetric spatial wave function $\psi_+(\mathbf{r_1},\mathbf{r_2})=\psi_+(\mathbf{r_2},\mathbf{r_1})$, and an antisymmetric spatial wave function $\psi_-(\mathbf{r_1},\mathbf{r_2})=-\psi_-(\mathbf{r_2},\mathbf{r_1})$. By this, it makes sense that when $\psi_a=\psi_b$, fermionic systems have no sensical spatial wave function (Pauli's exclusion principle).
Now, as Griffiths likes to do to simplify explanations, he leaves spin out of wave functions. One paragraph later, he shows how fermions are expected to be further away from each other than distinguishable particles, and the converse for bosons ("exchange interaction"): this uses only integrals over space, so I assume it's fine to generalise the result to wave functions including spin. If I've interpreted his text later on in the chapter correctly, we can conclude such behaviour merely based on spatial wave functions, so I'll call particles that repel like fermions, and equivalently have can be given a combined spatial wavefunction $\psi_-$, "spatially fermionic".
Here's the problem. He adds spin into the discussion of two-electron systems as a spinor factor $\chi(1,2)$, and asserts:
It is the whole [$\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$], not just the spatial part, that has to be antisymmetric with respect to exchange. (...) Thus the Pauli principle allows two electrons in a given position state, as long as their spins are in the singlet configuration.
This statement confuses me.
For one: does "not just" imply that fermions still need to be spatially fermionic, as was asserted when spin was not yet included in the discussion, or that only $\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$ need be antisymmetric?
Secondly: is "$\psi(\mathbf{r_1},\mathbf{r_2})$" an elementary function $\psi(\mathbf{r_1},\mathbf{r_2})=\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})$, or is it an artificially (anti)symmetrised wavefunction like $\psi_+(\mathbf{r_1},\mathbf{r_2})$ and $\psi_-(\mathbf{r_1},\mathbf{r_2})$? If it's the former, that would mean that the spatial factor $\psi(\mathbf{r_1},\mathbf{r_2})$ in the combined wave function for our two-fermion system $\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$ cannot whatsoever be treated equally to the artificially (anti)symmetrised $\psi_\pm(\mathbf{r_1},\mathbf{r_2})$. So, if we can't, and if we assume the answer to question 1 is that the system has to be spatial fermionic as well, then how will we (or nature) ever ensure that $\psi$ is properly (anti)symmetrised?
Thirdly: since $\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$ must just be antisymmetric, why can't we take the triplet configuration of the two electrons (which gives a symmetric $\chi(1,2)$), and have an antisymmetric spatial wavefunction $\psi(\mathbf{r_1},\mathbf{r_2})$? (This thread tries to answer, but I don't think it gives proper closure.)
Note to future readers regarding the third question:
After some discussion in the comments of the accepted answer, and repeatedly having studied the above quote in the context of the chapter again, I came to the correct interpretation of what exactly Griffiths tried to exclude when writing "the Pauli principle allows two electrons in a given position state, as long as their spins are in the singlet configuration".
His claim can be phrased as follows:
If $\Psi=\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$, then there exists no mathematical function $\psi(\mathbf{r_1},\mathbf{r_2})$ that is antisymmetric w.r.t. interchange of $\mathbf{r_1}$ and $\mathbf{r_2}$ and uses only one state $\psi_a$ instead of a $\psi_a$ and a $\psi_b$ (if you will, $\psi_a = \psi_b$).
In the accepted answer by ZeroTheHero, you'll find the explanation as to why this is true $-$ the essence is that antisymmetrisation happens through determinants in permutation group theory, and that those become 0 when any $\psi_a = \psi_b$.
The main consequence is, in the end, as stated at first: two identical fermions, e.g. electrons, can't occupy the same $\psi_a = \psi_b$ unless and only unless being in an antisymmetric, i.e. singlet, spin configuration, exactly because there exists no separable antisymmetric spatial wave function that would allow for a symmetric, i.e. triplet, spin configuration.
Additionally, after walking through the chapter once more with this claim in mind, it became apparent that my concept of "spatial fermionicity" is indeed a separate property two particles can have. In the accepted answer, it is established that two fermions (e.g. electrons) needn't be spatially fermionic for them to be fermions. However, the system can still have said property, or even its exact opposite: in paragraph 5.2.1 on excited helium states, it is discussed that in parahelium, the electrons are specifically "spatially bosonic" (their expected separation is smaller than for distinguishable particles), making them interact at a closer range on average, measurable in the higher energy for such states.