= Solution
Use the prepoint prescription for the separable <Hamiltonian>. Its regulated exponent is
$$
\sum_{i=1}^N\left[ip_i(q_{i+1}-q_i)-\Delta t_i\left(\frac{p_i^2}{2m}+V(q_i)\right)\right].
$$
Each momentum integration is an ordinary <Gaussian integral>:
$$
\int_{-\infty}^\infty dp_i\,e^{-\Delta t_i p_i^2/(2m)+ip_i(q_{i+1}-q_i)}
=\sqrt{\frac{2\pi m}{\Delta t_i}}\exp\left[-\frac{m(q_{i+1}-q_i)^2}{2\Delta t_i}\right].
$$
Therefore the measure written in the question gives exactly
$$
\boxed{\mathcal Dq_{\mathrm{raw},N}=\prod_{i=1}^N\sqrt{\frac{2\pi m}{\Delta t_i}}\,dq_i,\qquad
S_{E,N}=\sum_{i=1}^N\left[\frac{m(q_{i+1}-q_i)^2}{2\Delta t_i}+\Delta t_i V(q_i)\right].}
$$
The resulting configuration integral is $\int\mathcal Dq_{\mathrm{raw},N}e^{-S_{E,N}}$. This is <Gaussian momentum integration in a phase-space path integral>. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.
For a normalized <quantum-mechanical propagator>, Fourier completeness uses $dp_i/(2\pi)$ rather than $dp_i$. With that normalization the configuration measure is
$$
\boxed{\mathcal Dq_N=\prod_{i=1}^N\sqrt{\frac{m}{2\pi\Delta t_i}}\,dq_i.}
$$
This differs from the raw measure by $(2\pi)^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\ldots,q_N$ are integrated for propagation from an initial <wavefunction>. If both endpoints are fixed, omit $dq_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.
We now address the unheaded real-time continuation. Set $m=\hbar=1$. The <normalized short-time Schrödinger kernel> gives the final-slice recurrence
$$
\psi(q,t+\Delta t)=\frac1{\sqrt{2\pi i\Delta t}}\int_{-\infty}^{\infty}dq'\,
\exp\left[\frac{i(q-q')^2}{2\Delta t}-i\Delta t V(q')\right]\psi(q',t).
$$
The square-root branch is fixed by the usual damped <Fresnel integral>, or continuation from the Euclidean kernel. The hint's $dq'/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i)^{-1/2}$ fixes the identity limit and is included at every step.
Put $\eta=q'-q$. The normalized oscillatory Gaussian has moments
$$
\langle1\rangle=1,\qquad\langle\eta\rangle=0,\qquad\langle\eta^2\rangle=i\Delta t,\qquad\langle\eta^4\rangle=3(i\Delta t)^2.
$$
Taylor-expand the smooth <wavefunction> and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order $(\Delta t)^2$. Thus
$$
\psi(q,t+\Delta t)=\psi(q,t)+\frac{i\Delta t}{2}\partial_q^2\psi(q,t)-i\Delta t V(q)\psi(q,t)+O((\Delta t)^2).
$$
Subtract the initial value, divide by $\Delta t$ and take the limit:
$$
\boxed{i\partial_t\psi(q,t)=\left[-\frac12\partial_q^2+V(q)\right]\psi(q,t).}
$$
This proves the <Time-dependent Schrodinger equation> by the <Schrödinger equation from a short-time path integral> argument. Nonuniform partitions give the same limit when their largest time step tends to zero.
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