= Solution
Set $q=y-x$ and $z_t=x+tq$. The definition of <pushforward measure> gives the <confinement energy> $\mathcal V[\mu_t]=\int V(z_t)\,d\Pi$. The <chain rule> and <differentiation under the integral sign>, whenever justified, give
$$
\frac d{dt}\mathcal V[\mu_t]=\int\nabla V(z_t)\cdot q\,d\Pi,\qquad\boxed{\frac{d^2}{dt^2}\mathcal V[\mu_t]=\int q^TD^2V(z_t)q\,d\Pi.}
$$
For example, bounded <Hessian matrix> is sufficient: it gives $|\nabla V(z)|\leq C(1+|z|)$, and the first two differentiated integrands are bounded by $C'(1+|x|^2+|y|^2)$. The <dominated convergence theorem> applies because $\Pi$ has finite second <absolute moments>. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.
\b[The printed growth assumption alone does not ensure a classical second derivative.] Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative <smooth function>
$$
V(x)=1-\cos(2\pi x^3),\qquad p_n=cn^{-4}\quad(n\geq1),\qquad c^{-1}=\sum_{n\geq1}n^{-4},
$$
and the <probability measures> $\mu=\sum_np_n\delta_{n-1/2}$, $\nu=\sum_np_n\delta_{n+1/2}$. Both have finite second <absolute moments>. Their translation <transport plan> is optimal: its cost is one, while every <coupling> has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal V[\mu_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the <Fatou lemma> gives
$$
\liminf_{h\to0}\frac{F(t_0+h)+F(t_0-h)-2F(t_0)}{h^2}\geq\sum_np_nV''(n)=36\pi^2c\sum_n1=\infty.
$$
Thus the requested classical identity requires an additional integrability assumption on the <Hessian matrix>; quadratic growth controls the energy but not its <second derivative>.
For a <confinement potential> $V$ that is a $\lambda$-<strongly convex function>, the derivative formula, where valid, yields
$$
\boxed{\frac{d^2}{dt^2}\mathcal V[\mu_t]\geq\lambda\int|q|^2\,d\Pi=\lambda D^2.}
$$
The underlying <displacement convexity> conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply <convexity> of $V(x)-\lambda|x|^2/2$ pointwise and integrate:
$$
\boxed{\mathcal V[\mu_t]\leq(1-t)\mathcal V[\mu]+t\mathcal V[\nu]-\frac\lambda2t(1-t)D^2.}
$$
The same argument on every subinterval makes $t\mapsto\mathcal V[\mu_t]-\lambda D^2t^2/2$ <convex>, so the second derivative lower bound also holds in the sense of <distributions>.
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