Edit: my original answer was wrong because I wrongfully treated $\mathbb E^{B_s}[\ \cdot\ ]$ as a conditional expectation with respect to $B_s$, which it is not. It helps intuition to think of it in that way, but it is defined differently (see Øksendal under definition 7.1.1 for details).
Yes, the second line is indeed a consequence of the Markov property, but not in the way you wrote it. Here is a possible way to look at it:
First, note that the stochastic process given by $X_t := \left(B_t,\int_0^t V(B_s)\ ds\right)^T$ is an $\mathbb R^{d+1}$-valued Itô diffusion. Indeed, we have
$$ dX_t = \big(dB_t,\ V(B_t)dt\big)^T = \sigma(X_t)dB_t + b(X_t)dt $$
where, for all $x\in\mathbb R^d$, $b(x_1,\ldots,x_{d+1}) = (\mathbf 0_d,V(x_1,\ldots,x_d))^T$ and $\sigma(x)$ is the $(d+1)\times d$ matrix whose first $d$ lines are the $d\times d$ identity matrix, and the last line is full of $0$'s.
I claim that we have the following equality
Lemma: for all $s,t\ge 0$,
$$\mathbb E^{B_s}\left[e^{-\int_0^t V(B_u) dB_u}f(B_t)\mid \mathscr F_s\right] = e^{-\int_s^{t+s} V(B_u) dB_u}f(B_{t+s}) \tag1$$
Indeed, write for all $t, s\ge 0$:
$$X_{t+s} = X_s + \int_s^{t+s} b(X_u)du +\int_s^{t+s} \sigma(X_u)dBu $$
and note that the increment $X_{t+s}-X_s$ is independent from $\mathscr F_s \equiv \{\sigma(B_u),0\le u \le s\}$ (see here for a proof). Furthermore, by time homogeneity, we have
$$X_{t+s}-X_s = X_t - X_0\ \ \text{ in distribution} $$
(see chapter 7 of Øksendal's book for a proof).
Now we can piece everything together. Define $g:(x_1,\ldots,x_{d+1})\mapsto e^{-x_{d+1}}\cdot f(x_1,\ldots,x_{d})$, and note that
$$\begin{align}
\mathbb E^{B_s}\left[e^{-\int_0^t V(B_u) dB_u}f(B_t)\mid \mathscr F_s\right] &= \mathbb E^{B_s}\left[g(X_t)\mid \mathscr F_s\right] \ (\text{by def. of } g) \\
&=\mathbb E^{B_s}\left[g(X_{t+s} - X_s + X_0)\mid \mathscr F_s\right] \ (\text{same distribution})\\
&= \mathbb E^{B_s}\left[g(X_{t+s} - X_s + X_0)\right] \ (\text{independence from }\mathscr F_s)\\
&= \mathbb E^{B_s}\left[e^{-\int_s^{t+s} V(B_u) dB_u}f(B_{t+s} - B_s + B_0)\right] \ (\text{replacing})\\
&= e^{-\int_s^{t+s} V(B_u) dB_u}f(B_{t+s})
\end{align} $$
where the last equality is due to the fact that $B_0 = B_s$ under $\mathbb Q^{B_s}$. This concludes the proof of Lemma $(1)$ and implies the second inequality in the OP.
Finally, to prove that $\mathbb E^x\left[\mathbb{E}^x[Z_{s+t}f(B_{s+t})\mid\mathscr F_s]\right] = \mathbb{E}^x[Z_{s+t}f(B_{s+t})] $, remember that we have the tower property of conditional expectations
Tower property: If $X$ is integrable, and $\mathscr G$ and $\mathscr H$ are two sigma-algebras such that $\mathscr H\subseteq\mathscr G$, then
$$\mathbb E[\mathbb E[X\mid\mathscr G]\mid\mathscr H] = E[X\mid\mathscr H] $$
and apply it to $X\equiv Z_{s+t}f(B_{s+t})$, $\mathscr G \equiv\mathscr F_s$ and $\mathscr H = \{\Omega,\emptyset\} $.
