I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider

$$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{(m-k/2)j^2+rj}}{ \partial q^a}(1)
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

Now $\frac{\partial^a q^{(m-k/2) j^2+rj}}{ \partial q^a}(1)$ is a polynomial of degree $2a$ in $j$. So we can write this in terms of

$$ F(b,c) = \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}j^b
\frac{ \partial ^{c} q^{(k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{c}}(1)$$

where $b+2c \leq 2i$.

We want to show that this vanishes for $i < n$ and evaluate it for $i=n$. For the first, it suffices to show that for all $b$ and $c$ with $b+2c < 2n$ we have:

$$F(b,c)=0$$

But these terms all show up in computing the partial derivatives for the sum

$$\sum\limits_{j = 0}^{2n} {( - 1)}^{ j}
q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}$$

$$ \frac{ \partial ^i \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}
q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}}{ \partial q^i} (1) = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{rj}}{ \partial q^a}(1)
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)
$$

$$ = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j} (rj)(rj-1) \dots (rj+1-a)
\frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

$$ = \sum_{i=0}^a \binom{a}{i} \sum_{b=0}^a[j^b] (rj)(rj-1) \dots (rj+1-a) F(b,i-a) $$

Now suppose by induction that $F(b,c)=0$ for $b+c < i $ for some fixed $i \leq 2n$. Then this sum simplifies to only include the terms where $b=a$. Here

$$[j^a] (rj)(rj-1) \dots (rj+1-a) = r^a$$

so we obtain

$$= \sum_{i=0}^a \binom{a}{i} r^a F(a,i-a) $$

If $i<2n$ then this polynomial is identically $0$ by Max's computation, so all the coefficients vanish, so $F(a,i-a)=0$. This verifies the induction step.

Hence $F(b,c)=0$ for $b+c<2n$, which includes $b+2c<2n$, so the first $n-1$ partial derivatives vanish.

What about the $n$th partial derivative? Well the only $F(b,c)$ with $b + 2c \leq 2n$ that we don't know is $0$ is $F(2n,0)$. So dropping all the other terms we get

$$\frac{ \partial^n f(n,m,r,k)}{\partial q^n} (1)= F(2n,0) [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) $$

Clearly

$$ [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) = [j^{2n}] ( (m-k/2)j^2+rj) ((m-k/2)j^2+rj-1) \dots ((m-k/2)j^2+rj+1-n) = (m-k/2)^{n}$$

So we get a constant depending only on $n$ times $(2k-m)^n$. To get the correct constant, you can compute it directly, or just note that it is sufficient to handle one special case for each $n$ and note that you did the $f(n,0,1,2)$ case in the comments to Max's answer.