TL;DR: OP's title question (v7) about instantons in the Minkowski signature is physically meaningless. It is an irrelevant mathematical detour run amok. The connection to physics/Nature is established via a Wick rotation of the full Euclidean path integral, not bits and pieces thereof. Within the Euclidean path integral, it is possible to consistently expand over Euclidean instantons, but it meaningless to Wick rotation the instanton picture to Minkowski signature.
In more details, let there be given a double-well potential
$$V(x)~=~\frac{1}{2}(x^2-a^2)^2. \tag{A}$$
The Minkowskian and Euclidean formulations are connected via a Wick rotation
$$ t^E e^{i\epsilon}~=~e^{i\frac{\pi}{2}} t^M e^{-i\epsilon}.\tag{B} $$
We have included Feynman's $i\epsilon$-prescription in order to help convergence and avoid branch cuts and singularities. See also this related Phys.SE post.
I) On one hand, the Euclidean partition function/path integral is
$$\begin{align} Z^E~=~&Z(\Delta t^E e^{i\epsilon})\cr
~=~& \langle x_f | \exp\left[-\frac{H \Delta t^E e^{i\epsilon}}{\hbar}\right] | x_i \rangle \cr
~=~&N \int [dx] \exp\left[-\frac{S^E[x]}{\hbar} \right],\end{align}\tag{C} $$
with Euclidean action
$$\begin{align} S^E[x]
~=~&\int_{t^E_i}^{t^E_f} \! dt^E \left[ \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\right)^2+e^{i\epsilon}V(x)\right]\cr
~=~& \int_{t^E_i}^{t^E_f} \! dt^E \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}\right)^2 \cr
&\pm \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)}.\end{align}\tag{D}$$
and real regular kink/anti-kink solution$^1$
$$\begin{align} \frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}~\approx~&0 \cr
~\Updownarrow~ & \cr
x(t^E)
~\approx~&\pm a\tanh(e^{i\epsilon}\Delta t^E). \end{align}\tag{E}$$
Note that a priori space $x$ and time $t^E$ are real coordinates in the path integral (C). To evaluate the Euclidean path integral (C) via the method of steepest descent, we need not complexify space nor time. We are already integrating in the direction of steepest descent!
II) On the other hand, the corresponding Minkowskian partition function/path integral is
$$\begin{align} Z^M~=~&Z(i \Delta t^M e^{-i\epsilon})\cr
~=~& \langle x_f | \exp\left[-\frac{iH \Delta t^M e^{-i\epsilon}}{\hbar} \right] | x_i \rangle\cr
~=~& N \int [dx] \exp\left[\frac{iS^M[x]}{\hbar} \right],\end{align}\tag{F} $$
with Minkowskian action
$$\begin{align} S^M[x]~=~&\int_{t^M_i}^{t^M_f} \! dt^M \left[ \frac{e^{i\epsilon}}{2} \left(\frac{dx}{dt^M}\right)^2-e^{-i\epsilon}V(x)\right]\cr
~=~& \int_{t^M_i}^{t^M_f} \! dt^M \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}\right)^2 \cr
&\pm i \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)},\end{align}\tag{G}$$
and imaginary singular kink/anti-kink solution
$$\begin{align} \frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}~\approx~&0 \cr
~\Updownarrow~ & \cr
x(t^M)~\approx~&\pm i a\tan(e^{-i\epsilon}\Delta t^M)\cr
~=~&\pm a\tanh(i e^{-i\epsilon}\Delta t^M). \end{align}\tag{H}$$
It is reassuring that the $i\epsilon$ regularization ensures that the particle starts and ends at the potential minima:
$$ \lim_{\Delta t^M\to \pm^{\prime}\infty} x(t^M)~=~(\pm a) (\pm^{\prime} 1). \tag{I}$$
Unfortunately, that seems to be just about the only nice thing about the solution (H). Note that a priori space $x$ and time $t^M$ are real coordinates in the path integral (F). We cannot directly apply the method of steepest descent to evaluate the Minkowski path integral. We need to deform the integration contour and/or complexify time and space in a consistent way. This is governed by Picard-Lefschetz theory & the Lefschetz thimble. In particular, the role of the imaginary singular kink/anti-kink solution (H) looses its importance, because we cannot expand perturbatively around it in any meaningful way.
--
$^1$ The explicit (hyperbolic) tangent solution (E) is an over-simplified toy solution. It obscures the dependence of finite initial (and final) time $t^E_i$ (and $t^E_f$), moduli parameters, and multi-instantons. We refer to the literature for details.
Best Answer
Let me first refer you to three references pedagogically treating Instantons in quantum mechanics: 1)Riccardo Rattazzi's lecture notes treating instantons in nonsupersymmetric quantum mechanics. In these notes the anharmonic oscillator model is elaborated with great detail 2) Philip Argyres lecture notes treating instantons in supersymmetric quantum mechanics. The model considered in these lecture notes is a simplified one dimensional version Witten's model in flat space 3) Salomonson and van-Holten's original article, where they elaborate in detail the same model treated by Argyres. This article can be used to fill the gaps in Argyres' notes.
Regarding the first question:
First, one must notice that if we take a classical solution in which the fermionic coordinates are nonvanishing, then both the bosonic and fermionic coordinates acquire Grassmann components. The bosonic coordinate becomes an even Grassmann number and the fermionic and odd one. The action $S$ itself becomes an even Grassmann number. This casts a difficulty in the interpretation of $e^{-S}$ as a tunneling rate between the degenerate vacuua.
Salomonson and van-Holten (like Witten) discard the fermions (i.e., substitute zero for the fermion's "classical field"). They justify this substitution by the requirement of keeping the action finite (which is a crucial requirement because $e^{-S}$ is proportional to the transition rate between the vacuua). However, Akulov and Duplij find the general solution for the same model with a nonvanishing fermionic coordinate. They find that the Grassmann number contribution to the action identically vanishes and the action is equal to the classical action with the fermions discarded. This partly justifies the discarding of the fermions (Further justification will be given in the fermionic zero mode discussion). In addition Akulov and Duplij find that unlike the action, the Grassmann number dependence does not generally vanish in the instanton topological charge; this contribution vanishes exactly only for potentials breaking supersymmetry which is reminiscent of the vanishing of the Witten's index when supersymmetry is spontaneously broken. Furthermore Akulov and Duplij extend their analysis to a non-supersymmetric model and find that in this case the Grassmann number contribution to the action does not vanish.
The treatment of zero modes:
The fermion and boson determinants excluding the zero modes exactly cancel. As explained in Argyres, zero modes in the fermionic sector cause the partition function to vanish. However, the correction to the ground state energy entails the insertion of a supersymmetry generator (equation 4.16 in Argyres), this insertion is what is necessary to make the fermionic path integral nonvanishing, since for Grassmann variable $\int d\psi_0 = 0$ while $\int \psi_0 d\psi_0 = 1$.
Now, if we adopt the Akulov and Duplij method, the contribution of the classical fermionic field to the path integral vanishes because as already mentioned, the action does not depend on the classical fermionic variables, thus there is no insertion in the classical component, thus its contribution vanishes.
The bosonic zero mode corresponds to the instanton's collective coordinate (moduli space). Geometrically, this coordinate is the central time $t_0$ of the classical kink solution; and the solution satisfies the equations of motion for all $t_0$ values. The correct evaluation of the path integral requires to perform the bosonic integration on the nonzero modes which is finite, then integrate over the moduli space which entails the integration over $t_0$.
The path integral even in this simple case is quite cumbersome and its explicit calculation is given in Salomonson and van-Holten.
For a rigorous and detailed computation of the instanton path integral for the Witten's model, please see Alice Rogers article.