How many times does $k$ occur in the composition of $n$? How many times does the number $k$ occur in the composition of $n$?
Composition of Integer 
In short, the difference between the partition of an integer and composition is the order of numbers. In partition, the order doesn't matter whereas, in composition, it does. That's why Partitions are sometimes called as ordered Compositions.
Example: $k$ = $1$ & $n$ = $5$
The composition of 5 are:
$5$ 
$4 + 1$
$3 + 2$
$3 + 1 + 1$
$2 + 3$
$2 + 2 + 1$
$2 + 1 + 2$
$2 + 1 + 1 + 1$
$1 + 4$
$1 + 3 + 1$
$1 + 2 + 2$
$1 + 2 + 1 + 1$
$1 + 1 + 3$
$1 + 1 + 2 + 1$
$1 + 1 + 1 + 2$
$1 + 1 + 1 + 1 + 1$
In all $1$ occurs $28$ times in the composition of $5$
Similarly is there any relation between $k$ and $n$ for all $n \geq 0$ & $k \leq n$

The generating function here is
$$G(z, u) = \sum_{q\ge 1} \left(uz^k - z^k + \frac{z}{1-z}\right)^q.$$
We then have that
$$\left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1}
\\ = \left. \sum_{q\ge 1}
q \left(uz^k - z^k + \frac{z}{1-z}\right)^{q-1} z^k\right|_{u=1}
= z^k \sum_{q\ge 1} q \left(\frac{z}{1-z}\right)^{q-1}
\\ = z^k \frac{1}{(1-z/(1-z))^2}
= z^k \frac{1-2z+z^2}{(1-2z)^2}.$$
Extracting coefficients we find for $n\ge k+2$
$$[z^{n-k}] \frac{1}{(1-2z)^2}
- 2[z^{n-k-1}] \frac{1}{(1-2z)^2}
+ [z^{n-k-2}] \frac{1}{(1-2z)^2}
\\ = 2^{n-k} (n-k+1) - 2^{n-k} (n-k) + 2^{n-k-2} (n-k-1)
\\ = 2^{n-k} + 2^{n-k-2} (n-k-1)
= 2^{n-k-2} (n-k+3).$$
We get for $n=k+1$
$$2^{n-k} (n-k+1) - 2^{n-k} (n-k) = 2$$
which corresponds to $(1,k)$ and $(k,1)$
and for $n=k$
$$2^{n-k} (n-k+1) = 1.$$
These only depend on $n-k$ as pointed out in the comments.
