This is quite hand-wavy but I think that's acceptable considering you want intuition. There is $1$ key gap in the result, but maybe someone can come along and fill that in nicely.
First the notation:
$N(n)$ is the number of complete compositions of $n$
$N(n,a)$ is the number of complete compositions of $n$ with $a$ as the largest term
$N(n,a,i)$ is the number of complete compositions of $n$ with $a$ as the largest term and only a single term $i$ that has to be at the end of the sequence.
For each of those, let $\tilde{N}$ denote the set of sequences that fit the description.
For example: $N(5)=8$, $N(5,1)=1$, $N(5,2)=7$, $N(5,2,2)=1$, $N(5,2,1)=1$, $\tilde{N}(5,1)=\{1+1+1+1+1\}$, $\tilde{N}(5,2,2)=\{1+1+1+2\}$, $\tilde{N}(5,2,1)=\{2+2+1\}$
Now we introduce $2$ key relations. The first one:
$$N(n)=\sum_{a=1}^{n}N(n,a)$$
This one should make sense. The total number of complete compositions of $n$, is the number of complete compositions with a maximum term of $1$ + the number of complete compositions with a maximum term of $2$ + etc.
The second one:
$$N(n,a)=\sum_{i=1}^{a}N(n-i,a)+N(n,a,i)$$
To see this, take an arbitrary composition $c\in \tilde{N}(n,a)$. Write $c=c_1+c_2+...+c_m$, and consider the composition $c'=c_1+c_2+...+c_{m-1}$. Now there are $2$ possibilities:
- $c'$ is a complete composition of $n-c_m$ with largest term $a$
- $c'$ is an incomplete composition of $n-c_m$
If $c'$ is a complete composition of $n-c_m$ with largest term $a$, then $c'\in\tilde{N}(n-c_m,a)$.
Otherwise, since $c$ was a complete composition of $n$ with largest term $a$ and $c'$ is not complete, $c_m$ must have been the only term with this value in $c$, so that removing it results in an incomplete composition. Therefore $c'\in \tilde{N}(n,a,c_m)$
You can do this reasoning the other way around as well to establish the claim.
Now we first have to agree that
$$\lim_{n\to \infty}\frac{\sum_{i=1}^{a}N(n,a,i)}{N(n,a)}=0$$
To see this, think about what compositions $\cup_{i=1}^{a}\tilde{N}(n,a,i)$ contains. This set contains all compositions that contain some term exactly once, where that term has to be the last term in the composition. It is 'intuitively clear' that this is an awfully specific subset of $\tilde{N}(n,a)$.
So that we can rewrite*:
$$N(n,a)\approx\sum_{i=1}^{a}N(n-i,a)$$
And now
$$
\begin{aligned}
N(n+1)
& =\sum_{a=1}^{n+1}N(n+1,a) \\
& \approx \sum_{a=1}^{n+1}\sum_{i=1}^{a}N(n-(i-1),a)\\
& = \sum_{a=1}^{n+1}\sum_{i=0}^{a-1}N(n-i,a)\\
& = \sum_{a=1}^{n}\sum_{i=0}^{a-1}N(n-i,a) + \sum_{i=0}^{n}N(n-i,n) \\
& = \sum_{a=1}^{n}[\sum_{i=1}^{a}[N(n-i,a)] + N(n,a) - N(n-a,a)]\\
& = \sum_{a=1}^{n}\sum_{i=1}^{a}N(n-i,a) + \sum_{a=1}^{n}N(n,a) - \sum_{a=1}^{n}N(n-a,a)\\
& \approx N(n) + N(n) - \sum_{a=1}^{n}N(n-a,a)\\
& \approx 2N(n)
\end{aligned}
$$
So now we established that $N(n)$ asymptotically will behave as $c2^n$ for some constant $c$. Now from the fact that the total number of compositions of $n$ equals $2^{n-1}$ we know that $c<\frac{1}{2}$. And from observation we know of course that $c=\frac{1}{4}$. However, I have not really been able to come up with an argument for why it is $\frac{1}{4}$. Maybe it makes sense to say that $c$ must be of the form $\frac{1}{2^m}$ because we do want $N(n)$ to be a whole number. So then all we have to do is it find a reasoning for why $N(n)>2^{n-3}$. So maybe someone who reads this will be able to complete this reasoning from here.
So this might seem like a useless result because we are still left with the question of why this constant is exactly $\frac{1}{4}$. However, with this reasoning we at least see why the ratio converges in the first place.
*A remarkable result of this approximation is that it turns out that $N(n,2)\approx\frac{1+\sqrt 5}{2}F_n$ where $F_n$ is the $n$th fibonacci number. And this approximation is very good (always only either $1$ or $2$ too large).
Best Answer
We can also map the compositions of n into odd parts to the compositions of n-1 into 1s and 2s (which is the same as the number of ways of dividing a 2x(n-1) rectangle into 2x1 rectangles) as follows:
Given a composition of n into odd parts, split each part into 0 or more 2s followed by a 1. Remove the final 1. You now have a composition of n-1 into 1s and 2s.
To go in reverse, given a composition of n-1 into 1s and 2s, add an extra 1 at the end and then group parts in sequences of 0 or more 2s that end in a 1. Sum each group of 2s followed by a 1 to get a composition of n into odd parts.
For example, the compositions of 5 into odd parts and 4 into 1s and 2s are related as follows:
1+1+1+1+1 <-> 1+1+1+1
1+1+3 <-> 1+1+2
1+3+1 <-> 1+2+1
3+1+1 <-> 2+1+1
5 <-> 2+2