It's not true. For a counterexample, take $\Omega = \{a,b,c\}$ to be a sample space with 3 points, $\mathcal{F} = 2^{\Omega}$, and $P(A) = \frac{1}{3} |A|$ to be the uniform probability measure assigning probability 1/3 to each outcome. Let's represent a random variable $X : \Omega \to \mathbb{R}$ as the ordered triple $(X(a), X(b), X(c))$.
Set $\mathcal{G}_1 = \{\Omega, \emptyset, \{a\}, \{b,c\}\}$, and $\mathcal{G}_2 = \{\Omega, \emptyset, \{b\}, \{a,c\}\}$. Let $X$ be the random variable $(1,2,3)$. Then one can directly compute $$\begin{align*} E[X \mid \mathcal{G}_1] &= (1, 2.5, 2.5) \\ E[X \mid \mathcal{G}_2] &= (2,2,2) \\ E[E[X \mid \mathcal{G}_1] \mid \mathcal{G}_2] &= (1.75,2.5,1.75) \\ E[E[X \mid \mathcal{G}_2] \mid \mathcal{G}_1] &= (2,2,2). \end{align*}$$
To prove the second claim, I assert the following: suppose $U,V,W$ are mutually independent random variables or vectors, and $Y$ is $\sigma(U,V)$-measurable and integrable. Then $E[Y \mid V,W] = E[Y \mid V]$.
Clearly $E[Y \mid V]$ is $\sigma(V,W)$ measurable, so it suffices to show $E[Y 1_A] = E[E[Z \mid V] 1_A]$ for any $A \in \sigma(V,W)$. Suppose first that $A$ is of the form $A = B \cap C$ for $B \in \sigma(V)$, $C \in \sigma(W)$. Then $$\begin{align*} E[E[Y \mid V] 1_A] &= E[E[Y\mid V] 1_B 1_C] \\ &= E[E[Y 1_B \mid V] 1_C] && \text{since $B \in \sigma(V)$} \\ &= E[E[Y 1_B \mid V]] E[1_C] && \text{since $1_C \in \sigma(W)$ is independent of $\sigma(V)$} \\ &= E[Y 1_B] E[1_C] \end{align*}$$ while $E[Y 1_A] = E[(Y 1_B) 1_C] = E[Y 1_B] E[1_C]$ since $Y 1_B \in \sigma(U,V)$ is independent of $\sigma(W)$. The general case follows by a monotone class or $\pi$-$\lambda$ argument (the collection of all $B \cap C$ is a $\pi$-system that generates $\sigma(V,W)$).
Now apply this taking $U = (X_{i+1}, \dots, X_{n})$, $V = (X_{1}, \dots, X_{i-1})$ and $W = X_i$, and $Y = E[Z \mid U,V]$.