Let's start with the long Weierstrass equation for an elliptic curve over a field $K$:
$$ y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6$$
If the characteristic of the $K$ is not 2 then we can make the substitution $y_1=\dfrac{a_1}{2}(y-a_1x-a_3)$ which leads us to a medium equation:
$$y_1^2 = 4x^3 + b_2x^2 + 2b_4x+b_6$$
If the characteristic is not 3 then we can make another substitution $x_1=\dfrac{1}{36}(x-3b_2), y_2=\dfrac{y_1}{108}$ to get a short equation:
$$y^2=x^3-27c_4x-54c_6$$
If however, the characteristic of $K$ equals $2$ or $3$ then we can not do this, but you can instead show that it can be reduced to one of the four forms in your link.
Assume the characteristic is $3$. Then we can eliminate the $a_1$ and $a_3$ as above to get to a medium form. (Note we can take $y=2y'$ then divide through by $4$ to get rid of the coefficient in front of $x^3$. Now if $b_2=0$ (happens if and only if the j-invariant is $0$) then we are in the supersingular case so we are done.
Otherwise take $x=x'+\dfrac{2b_4}{b_2}$ to get something in the regular case.
Now assume the characteristic is $2$ and we start with the long equation. If $a_1=0$ (iff the j-invariant is 0 so we are in the supersingular case again), then take $x=x'+a_2$.
Otherwise, take $x=a_1^2x'+\dfrac{a_3}{a_1}$ and $y=a_1^3y' + \dfrac{a_1^2a_4+a_3^2}{a_1^3}$ to get something in the regular case.
