The first relationship is actually more interesting than you suggest. Specifically, applying your recipe to the ordered partitions of $n$ produces $F_{2n-1}$, while adding the products without reducing the last entry to $1$ produces $F_{2n}$.
It’s not too hard to see why this works. Suppose that it’s true for some $n$, and consider the ordered partitions of $n+1$. They are of two types: those that end in $1$, and those that don’t. Each partition of $n+1$ that ends in $1$ is obtained from a partition $\pi$ of $n$ by appending $+1$; its reduced product is the product of the entries in $\pi$, so the sum of the reduced products of these partitions of $n+1$ is $F_{2n}$.
Each partition of $n+1$ that does not end in $1$ is obtained from a partition $\pi$ of $n$ by adding $1$ to its last element; its reduced product is the same as the reduced product of $\pi$, so the sum of the reduced products of these partitions of $n+1$ is $F_{2n-1}$. Thus, the sum of the reduced products of all of the ordered partitions of $n+1$ is $F_{2n}+F_{2n-1}=F_{2n+1}=F_{2(n+1)-1}$, as desired.
The sum of the unreduced products of the partitions of $n+1$ that end in $1$ is the same as the sum of their reduced products, or $F_{2n}$, so to complete the argument, we need only show that the sum of the unreduced products of the partitions that do not end in $1$ is $F_{2n+1}$. But this is clear: if \pi' is an ordered partition of $n+1$ obtained by adding $1$ to the last element of some ordered partition $\pi$ of $n$, the unreduced product of \pi' is the sum of the reduced and unreduced products of $\pi$. Summing over all such partitions of $n+1$ then yields a total of $F_{2n}+F_{2n-1}=$ $F_{2n+1}$, just as in the last paragraph.
Of course the induction gets off the ground with no difficulty, since the sums for $n=1$ are $F_1=1=F_2$.
I’d meant to add this a while back, but I got busy and forgot:
This differs from savicko1’s argument chiefly in that it looks only at the immediately preceding integer.
The second question can be dealt with similarly. Let $P(n)$ be the set of ordered partitions of $n$. Call an ordered partition of $n$ odd or even according as the number of terms is odd or even. Let $P_o(n)$ be the set of odd ordered partitions of $n$, $P_o^-$ the set of odd ordered partitions of $n$ whose last term is $1$, and $P_o^+(n)$ be the set of odd ordered partitions of $n$ whose last term is greater than $1$, and define $P_e(n)$, $P_e^-(n)$, and $P_e^+(n)$ similarly for even ordered partitions of $n$. For each ordered partition $\pi$ of $n$ let $f(\pi)$ be the product of the factors $2^{x-1}$ as $x$ ranges over the odd-numbered terms of $\pi$. Finally, let $s(n)=\sum_{\pi\in P(n)}f(\pi)\text{ and }t(n)=\sum_{\pi\in P_o(n)}f(\pi)\;.$
Then $\begin{align*} &\sum_{\pi\in P_o^-(n+1)}f(\pi)=\sum_{\pi\in P_e(n)}f(\pi)=s(n)-t(n)\;,\\ &\sum_{\pi\in P_o^+(n+1)}f(\pi)=2\sum_{\pi\in P_o(n)}f(\pi)=2t(n)\;,\\ &\sum_{\pi\in P_e^-(n+1)}f(\pi)=\sum_{\pi\in P_o(n)}f(\pi)=t(n)\;,\text{ and}\\ &\sum_{\pi\in P_e^+(n+1)}f(\pi)=\sum_{\pi\in P_e(n)}f(\pi)=s(n)-t(n)\;, \end{align*}\tag{1}$
and since the lefthand sides of $(1)$ sum to $s(n+1)$, $ s(n+1)=2s(n)+t(n)\tag{2}$ and $t(n+1)=\sum_{\pi\in P_o^-(n+1)}f(\pi)+\sum_{\pi\in P_o^+(n+1)}f(\pi)=s(n)+t(n)\;.$ If $s(k)=F_{2k}$ for $k\le n$, then $\begin{align*} s(n+1)&=2F_{2n}+t(n)\\ &=2F_{2n}+s(n-1)+t(n-1)\\ &=2F_{2n}+s(n-1)+\big(s(n)-2s(n-1)\big)\qquad\text{(from }(2)\text{)}\\ &=3F_{2n}-F_{2n-2}\\ &=2F_{2n}+F_{2n-1}\\ &=F_{2n}+F_{2n+1}\\ &=F_{2n+2}\;, \end{align*}$
and the result follows by induction.