Fixing $q$ and $k$ we ask about the asymptotics of $[z^m] F(z)$ where
$$F(z) = \frac{1-z^k}{1-qz+(q-1)z^{k+1}}.$$
We will suppose that $m\ge k$ since the answer is $q^m$ otherwise.
First verify that the poles $\rho$ are all simple. We get with $\rho$
a root of the denominator
$$\lim_{z\to\rho} \frac{z-\rho}{1-qz+(q-1)z^{k+1}}
= \lim_{z\to\rho} \frac{1}{-q + (q-1)(k+1)z^{k}}.$$
Now we have
$$-q + (q-1)(k+1)\rho^{k} = -q + (q-1)(k+1)\rho^{k+1}/\rho
\\ = -q + (k+1)(q\rho-1)/\rho.$$
If this were zero we would get
$$\rho = \frac{k+1}{kq}.$$
which is rational. Applying the rational root theorem to the
denominator of $F(z)$ we must have
$$\frac{k+1}{kq} = \frac{1}{r}$$
where $r|(q-1).$ This requires $q=p(k+1)$ where $kp|(q-1).$ There are
two cases. The first is $kp=q-1$ which implies $q=q-1+p$ i.e. $p=1.$
We get $\rho = 1/k$ however for $1-(k+1)/k + k / k^{k+1} = 0,$ we
would need $1/k^k = 1/k$ so no contribution from this case (here we
have assumed that $k\ge 2$ since the answer is $(q-1)^m$ trivially
otherwise). Second case is that $kp$ is a proper divisor of $q-1$
which gives the bound $q\le (q-1)/2 + (q-1)/4$, a contradiction.
Hence all poles are simple.
Working with
$$G(z) = \frac{1}{1-qz+(q-1)z^{k+1}}.$$
we have under these conditions
$$G(z) = \sum_\rho \frac{1}{z-\rho} \mathrm{Res}_{z=\rho} G(z)
= - \sum_\rho \frac{1}{\rho} \frac{1}{1-z/\rho}
\mathrm{Res}_{z=\rho} G(z).$$
This last expression tells us that we need the pole closest to the
origin for the asymptotics. Observe that with $|z|\lt 1$ the term
$1-qz+(q-1)z^{k+1}$ is dominated by $1-qz$ and hence there must be a
pole in the vicinity of $z=1/q$, which it will require numerics to
approximate (e.g. start Newton's method at $1/q$) and which we will
call $\beta.$ The complex poles come in pairs so if we can show that
it is real and that there are no others with $|z|<1/q$ we have that
the contribution from $\beta$ yields the asymptotics. At this time we
will show that there is one zero from the denominator inside
$|z|=(1/q)\times q/(q-1) = 1/(q-1)$ by Rouche's Theorem, taking $f(z)
= 1-qz+(q-1)z^{k+1}$ and $g(z) = - qz$ where
$$|f(z)-g(z)| =
|1+(q-1)z^{k+1}| \le 1 + \frac{1}{(q-1)^k}
\quad\text{and}\quad |g(z)| = \frac{q}{q-1}
= 1 + \frac{1}{q-1}.$$
The conditions for the theorem hold (i.e. $|f(z)-g(z)| \lt |g(z)|$ on
$|z|=1/(q-1)$) when $k\ge 2$ which we assumed so $f(z)$ has the same
number of zeros inside the contour as $g(z)$ and $g(z)$ has just one
root there, so $f(z)$ does as well and this is our pole $\beta.$ As
there is just one it must be real.
To conclude here it remains to extract coefficients which yields
$$-\frac{1}{\beta}
\left(\frac{1}{\beta^m}-\frac{1}{\beta^{m-k}}\right)
\times \frac{1}{-q+(k+1)(q\beta-1)/\beta}.$$
We factorize to obtain the constant and find
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{(k+1)(q\beta-1)-q\beta}
\left(\beta^k-1\right)
\times \frac{1}{\beta^{m}}.}$$
This formula was verified numerically where $\beta$ was computed and
the evidence indicates rapid convergence with $m.$
Addendum. To help OP we also include the closed form which
requires
$$[z^m] \frac{1}{1-qz+(q-1)z^{k+1}}
= [z^m] \sum_{p\ge 0} z^p (q-(q-1) z^k)^p
\\ = \sum_{p=0}^m [z^{m-p}] (q-(q-1) z^k)^p
= \sum_{p=0}^m [z^p] (q-(q-1) z^k)^{m-p}
\\ = \sum_{p=0}^{\lfloor m/k\rfloor} [z^{kp}] (q-(q-1) z^k)^{m-kp}.$$
Here we must have $m-kp\ge p$ and we find
$$[z^m] G(z) = g_m = \sum_{p=0}^{\lfloor m/(k+1)\rfloor}
{m-kp\choose p} (1-q)^p q^{m-(k+1)p}$$
and the closed form is given by $f_m = [z^m] F(z) = g_m-g_{m-k}.$
