How can we produce another geek clock with a different pair of numbers?

For $n=12$ and $k=12$ here is a solution:

$1=\frac{12}{12+12+12+12+12+12-(12+12+12+12+12)}$

$2=\left(12 \times \frac{12}{12-12+12-12+12+12+12+12+12+12}\right)$

$3=\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$

$4=\left(12-\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$5=\left(12 \times \frac{12}{\left(12 \times \left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$6=\left(12+\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$7=\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$8=\left(12+\left(12 \times \frac{12}{\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)$

$9=\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$10=\left(12 \times \frac{12}{\left(12-\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$11=\left(12+\frac{12}{\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)}\right)$

$12=\left(12+\left(12+\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)\right)\right)$

Making numbers out of 4 fours is a common problem: $$1=\frac {44}{44}$$ $$2=\frac {4\cdot 4}{4+4}$$ $$3=\frac{4+4+4}{4}$$ $$4=\frac{4-4}{4}+4$$ $$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$ $$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$ $$7=\sqrt{4!\sqrt 4+\frac 4 4}$$ $$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$ $$9=(4-\frac 4 4)^{\sqrt 4}$$ $$10=\frac{4!} 4 - (4-\sqrt 4)$$ $$11=\frac{4!}{\sqrt 4}-\frac 4 4$$ $$12=\sqrt{\frac{4!4!}{\sqrt 4+\sqrt 4}}$$

You should clarify what operations you want. If you allow for any kind of rounding function, factorials and logs you can almost certainly do it with one of any number (though the resulting expressions may not fit on a clock).

solution for n = 1, k = 12:

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1 = 4 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1 = 5 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1 = 6 $$

$$ 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1 = 7 $$

$$ 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1 = 8 $$

$$ 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1+1 = 9 $$

$$ 1 \times 1 \times 1+1+1+1+1+1+1+1+1+1 = 10 $$

$$ 1 \times 1+1+1+1+1+1+1+1+1+1+1 = 11 $$

$$ 1+1+1+1+1+1+1+1+1+1+1+1 = 12 $$