I) Bosonic part: When we Wick-rotate, it is more natural to use signature convention $(-,+,+,+) $
for the Minkowski (M) metric $g^M_{\mu\nu}$, and $(+,+,+,+)$ for the Euclidean (E) metric $g^E_{\mu\nu}$, cf. e.g. my Phys.SE answer here. We will use Greek indices $\mu,\nu=0,1,2,3$, to denote curved spacetime indices; Roman indices $a,b=0,1,2,3$, for flat spacetime indices; and Roman indices $j,k=1,2,3$, for spatial indices. We shall not relabel $x^0=x^4$ to avoid messing with the orientation.
Standard conventions for the Wick rotation are [1]
$$\begin{align} -S_E~=~&iS_M \quad\text{action}, \cr
t_E~=~&it_M \quad\text{time}, \cr
-{\cal L}_E~=~&{\cal L}_M \quad\text{Lagrangian density}, \cr
d^4x_E~=~&id^4x_M \quad\text{spacetime $4$-form}, \cr
-\mathbb{L}_E~=~&i\mathbb{L}_M \quad\text{Lagrangian $4$-form}, \cr
V_E^0~=~&iV_M^0 \quad\text{zero-comp. of contravariant vector}, \cr
V^M_0~=~&iV^E_0 \quad\text{zero-comp. of covariant vector},
\end{align}\tag{1}$$
and a straightforward generalization to tensor fields $T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}$.
(NB: The indices of the Levi-Civita symbol are not Wick-rotated, as the values of the symbol only consists of $\pm 1$ and $0$.)
Both curved and flat indices are Wick-rotated. For this reason the determinant of the vielbein $$e~=~\det(e^a_{\mu})\quad\text{and}\quad\sqrt{|g|}~=~\sqrt{\det(g_{\mu\nu})\det(\eta^{ab})}\tag{2}$$
are invariant under Wick-rotation.
Example: Topological/theta/Chern-Simons term.
Consider a Lagrangian 4-form term of the form of a $4$-form
$$\begin{align} i\mathbb{L}_E~\stackrel{(1)}{=}~&\mathbb{L}_M~=~\Omega_M~=~d^4x_M ~\Omega^M_{0123}\cr
~\stackrel{(1)}{=}~&d^4x_E ~\Omega^E_{0123}~=~\Omega_E.
\end{align}\tag{3} $$
The corresponding Lagrangian density term reads
$$\begin{align} -{\cal L}_E~\stackrel{(1)}{=}~& {\cal L}_M~=~\frac{e}{4!}\varepsilon^{\mu_0\mu_1\mu_2\mu_3} \Omega^M_{\mu_0\mu_1\mu_2\mu_3}~=~ \Omega^M_{0123}\cr
~\stackrel{(1)}{=}~&i\Omega^E_{0123}~=~\frac{ie}{4!}\varepsilon^{\mu_0\mu_1\mu_2\mu_3} \Omega^E_{\mu_0\mu_1\mu_2\mu_3} .\end{align}\tag{4} $$
Example: Kinetic $F\wedge \star F$ term.
$${\cal L}_E~\stackrel{(1)}{=}~-{\cal L}_M~=~\frac{e}{4}F^M_{\mu\nu}F_M^{\mu\nu}~\stackrel{(1)}{=}~\frac{e}{4}F^E_{\mu\nu}F_E^{\mu\nu}.\tag{5}$$
Example: QED in flat space.
Let us here only consider QED (abelian gauge theory), and leave it to the reader to generalize to QCD (non-abelian gauge theory). The zero-component of the gauge variables (with indices down) is a co-vector/one-form and should transform like a time derivative
$$ \frac{\partial}{\partial t_M}~\stackrel{(1)}{=}~i \frac{\partial}{\partial t_E}\tag{6}$$
under Wick rotation. This implies
$$ -A^0_M~=~A^M_0~\stackrel{(1)}{=}~iA^E_0~=~iA^0_E, \qquad F^M_{0j}~\stackrel{(1)}{=}~iF^E_{0j},\tag{7}$$
Therefore the Maxwell Lagrangian density transforms as
$$ {\cal L}_M~=~-\frac{1}{4}F^M_{\mu\nu}F_M^{\mu\nu}~=~\frac{1}{2}F^M_{0j}F^M_{0j}-\frac{1}{4}F_{jk}F_{jk}, \tag{8}$$
$$ \begin{align} {\cal L}_M~=~&{\cal T}_M-{\cal V},\cr {\cal T}_M~=~&\frac{1}{2}F^M_{0j}F^M_{0j}, \cr {\cal V}~=~&\frac{1}{4}F_{jk}F_{jk};\end{align}\tag{9}$$
and
$$ {\cal L}_E~=~\frac{1}{4}F^E_{\mu\nu}F_E^{\mu\nu}~=~\frac{1}{2}F^E_{0j}F^E_{0j}+\frac{1}{4}F_{jk}F_{jk},\tag{10}$$
$$ \begin{align} {\cal L}_E~=~&{\cal T}_E+{\cal V},\cr {\cal T}_E~=~&\frac{1}{2}F^E_{0j}F^E_{0j}, \cr {\cal V}~=~&\frac{1}{4}F_{jk}F_{jk}.\end{align}\tag{11}$$
In particular, an Euclidean Lagrangian density ${\cal L}_E$ looks like a standard Lagrangian density (i.e. kinetic term minus potential term), with an apparent potential equal to minus ${\cal V}$.
II) Fermionic part: Wick rotation of spinor fields is a well-known non-trivial problem, cf. e.g. Refs. 2-4.
References:
W. Siegel, Fields; p. 329.
P. van Nieuwenhuizen and A. Waldron, A continuous Wick rotation for spinor fields and supersymmetry in Euclidean space, arXiv:hep-th/9611043.
A.J. Mountain, Wick rotation and supersymmetry, 1999.
A. Bilal & S. Metzger, arXiv:hep-th/0307152.
If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type
$$\int \frac{1}{k^n + ...} \cdot ... $$
Where $k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice that $(p \cdot k)_{+---}=-(p \cdot k)_{-+++}$ - by performing the Wick rotation in the inverse signature we will only get an overall minus sign and that does not change the applicability of our trick. So it poses no problem.
As you propose, as long as you take care of not crossing any singularities with your integration contour, you can perform a Wick rotation in other coordinates. If you do this properly without forgetting the factors from differentials or changed orientation of the contour, the result should be the same.
Do note that a number of other formulas also change sign when switching between signatures. E.g.
$$\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu \nu},\, (+---) \to \{\gamma^\mu,\gamma^\nu\} = -2 g^{\mu \nu},\,(-+++)$$
and so on. So to compare the results of different signatures, you would have to trace back all these signs and correct them. In the end, it really does not change any of the tricks of the calculations, let even the experimental predictions.
Best Answer
For the Euclidean path integral to be convergent, the Boltzmann factor should be an exponentially decaying function of the scalar field $\phi$. This in turn dictates that the direction of the Wick rotation. (It should perhaps be stressed that Wick rotation is not merely a renaming of time variables, but that it involves an actual deformation of the time-integration contour in the complex time plane.) Standard conventions for the Wick rotation are
$$\begin{align} -S_E~=~&iS_M, \cr t_E~=~&it_M, \cr {\cal L}_E~=~&-{\cal L}_M,\end{align} \tag{1} $$
where the subscripts $E$ and $M$ stand for Euclid and Minkowski, respectively. In OP's case $$\begin{align} {\cal L}_M~=~&i\phi^{\ast} \frac{d\phi}{dt_M}- {\cal V} ,\cr {\cal L}_E~=~&\phi^{\ast} \frac{d\phi}{dt_E}+ {\cal V} ,\cr {\cal V}~=~&\frac{1}{2m}\nabla\phi^{\ast}\cdot \nabla\phi .\end{align} \tag{2}$$ See also this related Phys.SE post.