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).
The generating function for distinct partitions into $k$ parts using elements from the set $\{1,\ldots,m\}$ is given using the $q$-binomial coefficient
$$\binom{m}{k}_q=\prod_{j=1}^{k}\frac{1-q^{m+1-j}}{1-q^j}\, .$$
The desired generating function is simply:
$$q^{\binom{k+1}{2}}\binom{m}{k}_q=\sum_{n=\binom{k+1}{2}}^{\binom{k+1}{2}+mk}p_{n,k,m}^*q^n\, ,$$
where $p_{n,k,m}^*$ is the number of partitions of $n$ in to $k$ distinct parts with no part greater than $m$.
We can use, for example, the free open source computer algebra system sage to expand
$$q^{\binom{6}{2}}\binom{10}{5}_q $$
which lists, as it's coefficients, partitions into distinct parts with $k=5$ parts and no part greater than $m=10$. Using the sage input:
j=var('j')
f(x)=x^15*product((1-x^(11-j))/(1-x^j),j,1,5)
show(expand(f(x)))
we get output:
$$\begin{align}&x^{40} + x^{39} + 2 \, x^{38} + 3 \, x^{37} + 5 \, x^{36} + 7 \, x^{35} + 9 \, x^{34} + 11 \, x^{33} + 14 \, x^{32} + 16 \, x^{31} + 18 \, x^{30}\\& + 19 \, x^{29} + 20 \, x^{28} + 20 \, x^{27} + 19 \, x^{26} + 18 \, x^{25} + 16 \, x^{24} + 14 \, x^{23} + 11 \, x^{22} + 9 \, x^{21} +\\& 7 \, x^{20} + 5 \, x^{19} + 3 \, x^{18} + 2 \, x^{17} + x^{16} + x^{15}\end{align}$$
Here we have 1 such partition for each of 15 and 16, 2 such partitions of 17, 3 partitions of 18 and so on.
The show command allows us to retrieve the output in $\LaTeX$ format so that, for example, it can be used to illustrate an answer on MSE ;)
By the way: A composition is different from a partition. For a composition the order of it's summands counts, whereas for partitions it doesn't.
Perhaps you already know this, however I am just making it clear for any other readers who may not be aware of the distinction. Of course, since we are counting partitions in to $k$ distinct parts then the number of compositions can be counted by multiplying by $k!$.
As per @GCab's request (see comments) I am posting a download link for more information on $q$-binomial coefficients:
A Course in Enumeration by Martin Aigner (pdf download)
The relevant section is 1.6.
Aigner gives a lucid account of the connection between $q$-binomial coefficients, lattice paths and restricted partitions. He then goes on to link, via a well known bijection, to the restricted partitions into distinct parts (corollary 1.2) talked about here.
Best Answer
Let the function $f(n, k, a, b, i)$ gives the $i$-th lexicographically smallest partition of $n$ into $k$ parts, each between $a$ and $b$, inclusively. I show how to compute $f$.
Let $g(n, k, a, b)$ be the number of partitions of $n$ into $k$ parts, each between $a$ and $b$, inclusively. Clearly $g(n, k, a, b) = \sum_{a \le j \le b} g(n-j, k-1, a, b)$.
To compute $f$, first we try all the possible first entry of the partition. For each possible entry $a \le j \le b$, we know that there are $g(n-j, k-1, a, b)$ partitions of $n$ with $j$ as its first element. Thus, we will be able to know what is the first entry of the $i$-th partition by finding the smallest $j$ such that $\sum_{a \le k \le j} g(n-j, k-1, a, b) \ge i$. The entire partition is then $j$ appended with $f(n-j, k-1, a, b, i - \sum_{a \le z \le j-1} g(n-z, k-1, a, b))$.