Lorentz spinors appear as irreducible representations of the group SL(2,C). Elements of the group are 2x2 matrices with complex entries and unity determinant. A Lorentz spinor is a two component vector $\psi^{A},\chi^{A}\in V_{2}$ with $A=1,2$. The Levi-Civita tensor $\epsilon_{AB}$ is an invariant tensor under SL(2,C). This means that if we have an irrep $\psi^{A}$ of SL(2,C) we can transform it with the Levi-Civita tensor and this will give an equivalent irrep. So, the covariant vectors $\psi_{A}\in \tilde{V}_{2}$ made as $\psi_{A}=\psi^{B}\epsilon_{BA}$ are equivalent to the contravariant vectors $\psi^{A}$; it doesn't matter whether we use $\psi^{A}$ or $\psi_{A}$ because both quantities represent the same physical thing.
The situation is the same as in Minkowski spacetime when we use $x^{\mu}$ or $x_{\mu}=\eta_{\mu\lambda}x^{\lambda}$. Covariant and contravariant vectors in Minkowski spacetime are equivalent irreps of the Lorentz group $O(1,3)$ because the metric $\eta_{\mu\lambda}$ is an invariant tensor.
The only problem with using the Levi-Civita tensor to lower a spinor index is that it is antisymmetric so that it matters which index is summed. I've decided to lower spinor indices as $\psi_{A}=\psi^{B}\epsilon_{BA}$ and so, for consistency, I've got to stick to this convention and not be tempted to use $\psi_{A}=\epsilon_{AB}\psi^{B}$.
Having picked a lowering convention, I'm forced to raise a spinor index with $\psi^{A}=\epsilon^{AB}\psi_{B}$ because then both operations are consistent.
\begin{equation}
\psi_{A}=\psi^{B}\epsilon_{BA}=\epsilon^{BC}\psi_{C}\epsilon_{BA}=\delta^{C}_{A}\psi_{C}=\psi_{A}
\end{equation}
Now we can make a SL(2,C) scalar $\psi_{A}\chi^{A}=\chi^{A}\psi_{A}$. The order of the vectors does not matter because the components $\psi^{1},\psi^{2}$ are just complex numbers. The following bit of index gymnastics recovers the property in the second equation in soliton's question.
\begin{equation}
\psi_{A}\chi^{A}=\psi^{B}\epsilon_{BA}\chi^{A}=-\psi^{B}\chi^{A}\epsilon_{AB}=-\psi^{A}\chi_{A}
\end{equation}
Everything so far has been classical. When we go over to quantum theory, the spinors are promoted to operators $\psi^{A}\rightarrow \hat{\psi}^{A}$. These operators represent fermions: as operators, they have to obey anticommutation relations, in this case $[\hat{\psi}^{A},\hat{\chi}^{B}]_{+}=0$. So, repeating the last calculation with operators,
\begin{equation}
\hat{\psi}_{A}\hat{\chi}^{A}=\hat{\psi}^{B}\epsilon_{BA}\hat{\chi}^{A}=-\hat{\psi}^{B}\hat{\chi}^{A}\epsilon_{AB}=-\hat{\psi}^{A}\hat{\chi}_{A}=+\hat{\chi}_{A}\hat{\psi}^{A}
\end{equation}
recovers the first equation in soliton's question.
I have to own up to the fact that I've not yet studied supersymmetry, but I think this is what must be going on based on general principles.
$\require{cancel}$
First of all, recall that a super-Lie bracket $[\cdot,\cdot]_{LB}$ (such as, e.g., a super-Poisson bracket $\{\cdot,\cdot\}$ & the super-commutator $[\cdot,\cdot]$), satisfies super-antisymmetry
$$ [f,g]_{LB} ~=~ -(-1)^{|f||g|}[g,f]_{LB},\tag{1} $$
and the super-Jacobi identity
$$\sum_{\text{cycl. }f,g,h} (-1)^{|f||h|}[[f,g]_{LB},h]_{LB}~=~0.\tag{2}$$
Here $|f|$ denotes the Grassmann-parity of the super-Lie algebra element $f$. Concerning supernumbers, see also e.g. this Phys.SE post and links therein.
In order to ensure that the Hilbert space has no negative norm states and that the vacuum state has no negative-energy excitations, the Dirac field should be quantized with anticommutation relations
$$ [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+}
~=~ \hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1}
~=~[\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+}, $$
$$ [\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)]_{+}
~=~ 0, \qquad [\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)]_{+}~=~ 0, \tag{3} $$
rather than with commutation relations, cf. e.g. Ref. 1 and this Phys.SE post.
According to the correspondence principle between quantum and classical physics, the supercommutator is $i\hbar$ times the super-Poisson bracket (up to possible higher $\hbar$-corrections), cf. e.g. this Phys.SE post. Therefore the corresponding fundamental super-Poisson brackets read$^1$
$$ \{\psi_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\}
~=~ -i\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})
~=~\{\psi^{\ast}_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\}, $$
$$ \{\psi_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\}
~=~ 0, \qquad \{\psi^{\ast}_{\alpha}({\bf x},t), \psi^{\ast}_{\beta}({\bf y},t)\}~=~ 0. \tag{4} $$
Comparing eqs. (1), (3) & (4), it becomes clear that the Dirac field is Grassmann-odd, both as an operator-valued quantum field $\hat{\psi}_{\alpha}$ and as a supernumber-valued classical field $\psi_{\alpha}$.
It is interesting that the free Dirac Lagrangian density$^2$
$$ {\cal L}~=~\bar{\psi}(\frac{i}{2}\stackrel{\leftrightarrow}{\cancel{\partial}} -m)\psi \tag{5} $$
is (i) real, and (ii) its Euler-Lagrange (EL) equation is the Dirac equation$^3$
$$(i\cancel{\partial} -m)\psi~\approx~0,\tag{6}$$
irrespectively of the Grassmann-parity of $\psi$!
The Dirac equation (6) itself is linear in $\psi$, and hence agnostic to the Grassmann-parity of $\psi$.
References:
M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 3.5.
H. Arodz & L. Hadasz, Lectures on Classical and Quantum Theory of Fields, Section 6.2.
--
$^1$ In this answer, we are for simplicity just considering dequantization, i.e. going from a quantum system to a classical system. Normally in physics, one is faced with the opposite problem: quantization. Given the Lagrangian density (5), one could (as a first step in quantization) find the Hamiltonian formulation via the Dirac-Bergmann recipe or the Faddeev-Jackiw method. The Dirac-Bergmann procedure leads to second class constraints. The resulting Dirac bracket becomes eq. (4). The Faddeev-Jackiw method leads to the same result (4). For more details, see also this Phys.SE post and links therein.
$^2$ The variables $\psi^{\ast}_{\alpha}$ and $\bar{\psi}_{\alpha}$ are not independent of $\psi_{\alpha}$, cf. this Phys.SE post and links therein. We disagree with the sentence "Let us stress that $\psi_{\alpha}$, $\bar{\psi}_{\alpha}$ are independent generating elements of a complex Grassmann algebra" in Ref. 2 on p. 130.
$^3$ Conventions. In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra
$$\{\gamma^{\mu}, \gamma^{\nu}\}_{+}~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}.\tag{7}$$
Moreover,
$$\bar{\psi}~=~\psi^{\dagger}\gamma^0, \qquad (\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{8} $$
The Hermitian adjoint of a product $\hat{A}\hat{B}$ of two operators $\hat{A}$ and $\hat{B}$ reverses the order, i.e.
$$(\hat{A}\hat{B})^{\dagger}~=~\hat{B}^{\dagger}\hat{A}^{\dagger}.\tag{9} $$
The complex conjugation of a product $zw$ of two supernumbers $z$ and $w$ reverses the order, i.e. $$(zw)^{\ast}~=~w^{\ast}z^{\ast}.\tag{10} $$
Best Answer
The classical limit requires many particles to be in the same state. This allowed us to experience bosonic fields like the spin-1 electromagnetic field or spin-2 gravitational field as classical limits of quantum fields.
Fermions, by definition, cannot have multiple particles in the same state, and so there is no "classical field" limit of a fermion.
The fermionic field in QFT plays a formal role in defining the quantum theory. You should treat it at as a tool we use to calculate quantum amplitudes for processes involving fermions. You should not try to interpret it classically as a field we could directly measure or that could be in a coherent state like a bosonic field.
As you said, in the path integral formalism, fermions are represented by Grassman-valued fields in the path integral. The fact that a field appears in the path integral, does not imply that there is a limit where we can directly experience a classical version of this field as a coherent state. Instead, doing the path integral with a Grassman field allows us to answer typical questions we want to ask of a quantum field theory, like what is the amplitude for a given scattering process to occur.
In the operator formalism, the fermionic field operator transforms as a spinor and obeys canonical anti-commutation relationships; it is not a Grassman function. Much like a scalar field, there is a difference between the mathematical object that represents the fermionic field in the operator formalism, and in the path integral formalism.