How does cutting a spring increase spring constant?

Equations are nice, but if you're looking for a conceptual answer :

In general, the spring constant or stiffness of a coiled spring is given as $$k=\frac{\pi Gd^4}{64R^3n}$$

Where, $G$ is modulus of rigidity of spring material

$d$ is the diameter of spring wire

$R$ is the mean radius of coil

$n$ is the effective number of coils in spring which is directly proportional to the length of coiled spring i.e. $n\propto L$

Above formula simply shows that if the other parameters are kept constant then $k$ is inversely proportional to $n$ number of coils. Since the number of coils $n$ is directly proportional to length $L$ of coiled spring hence its spring constant $k$ is inversely proportional to its length. Thus

$$k\propto \frac 1n\iff k\propto \frac1L$$ Therefore if a spring is broken into certain no. of pieces, all parameters $G$, $R$ & $d$ remain constant for all the pieces except the number of coils or turns $n$ decreases hence stiffness $k$ increases.

In general, if a coiled spring of length $L$ & stiffness $k$ is broken into $m$ no. of pieces of lengths $L_1, L_2, L_3, \ldots L_m$ then their respective spring constant or stiffness is given as follows $$k_1=k\left(\frac{L}{L_1}\right), \ \ \ k_2=k\left(\frac{L}{L_2}\right),\dots k_i=k\left(\frac{L}{L_i}\right), \ldots , k_m=k\left(\frac{L}{L_m}\right)$$
Relation among spring constants of broken pieces & original spring: $$\implies L1=\frac{kL}{k_1}, \ L2=\frac{kL}{k_2}, \ldots , Li=\frac{kL}{k_i}\ldots, Lm=\frac{kL}{k_m}$$ If we add all $m$ number of lengths of pieces we get original length $L$ of spring i.e. $$L_1+L_2+\ldots +L_i+\ldots +L_m=L$$ $$\frac{kL}{k_1}+\frac{kL}{k_2}+\ldots+\frac{kL}{k_i}+\ldots+\frac{kL}{k_m}=L$$ $$KL\left(\frac{1}{k_1}+\frac{1}{k_2}+\ldots+\frac{1}{k_i}+\ldots+\frac{1}{k_m}\right)=L$$ $$\color{blue}{\frac{1}{k_1}+\frac{1}{k_2}+\ldots+\frac{1}{k_i}+\ldots+\frac{1}{k_m}=\frac1k}$$

Above relation of spring constants is analogous to parallel connection of $m$ number of electric resistances

Let us consider that you join these n pieces of springs in series. Now you know that you have got the original spring whose spring constant is $k$ (say). Now joining springs in series is like joining resistors in parallel (identical formulae) which can easily proven by balancing forces. Hence,

$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'} + \frac{1}{k'}+\dots n ~\rm times$$ where $k'$ is the spring constant of individual cut springs.

On solving the above equation you will get that the spring constant becomes $n$ times.