Let's begin with a usual Artin-Schreier extension $k(x)/k$ with $F(x)-x=f$ where $F$ is the endomorphism of $k$ raising everything to power $p$, and $f\in k$ is not of the form $g^p-g$ for any $g\in k$. Here the automorphisms are gotten by extending $x\mapsto x+a$, $a\in \Bbb{F}_p$. The reason why these work is that $F(a)=a$, so not surprisingly $F(x+a)-(x+a)= F(x)-x$. The Galois group is then isomorphic to the additive group of the prime field, i.e. cyclic of order $p$.

To get cyclic extensions of degree $p^2$ we use the arithmetic of the ring of Witt vectors $W_2(k)$ of length $2$. The Frobenius map $F$ gives us (by functoriality of $W_2$) the endomorphism
$$W_2(F):W_2(k)\to W_2(k), (z_1,z_2)\mapsto (F(z_1),F(z_2))=(z_1^p,z_2^p).$$
The fixed points of $W_2(F)$ are obviously the elements of the subring $W_2(\Bbb{F}_p)$ that you undoubtedly know to be isomorphic to the ring $\Bbb{Z}/p^2\Bbb{Z}$. The way to turn this into a Galois group is to consider extensions $k(x_1,x_2)/k$, where the elements $x_1,x_2$ satisfy the Witt vector
equation
$$
W_2(F)(x_1,x_2)-(x_1,x_2)=(f_1,f_2)\qquad(*)
$$
for some suitable element $(f_1,f_2)\in W_2(k)$ (you need to exercise a bit of care to make sure that $(*)$ yields minimal polynomials of $x_1$ and $x_2$). This time the addition of any vector $(a_1,a_2)\in W_2(\Bbb{F}_p)$ gives an automorphism
$$(x_1,x_2)\mapsto (x_1,x_2)+(a_1,a_2).$$
This works because $(a_1,a_2)$ is a fixed point of $W_2(F)$ and therefore
$$
W_2(F)(x_1,x_2)-(x_1,x_2)=W_2(F)\big((x_1,x_2)+(a_1,a_2)\big)-\big((x_1,x_2)+(a_1,a_2)\big).
$$

May be this is not quite what you wanted to see? I did describe the cyclic extension, not as a simple step as you seem to want, but as a tower of
two cyclic extensions $k(x_1,x_2)/k(x_1)/k$.

So let me try and further illuminate this with a concrete example. I use $p=2$ because the Witt vector arithmetic doesn't look so unwieldy in that case. If $A$ is any commutative $\Bbb{F}_2$-algebra, the operations in $W_2(A)$ are given by
$$
\begin{aligned}
(a_1,a_2)+(b_1,b_2)&=(a_1+b_1,a_2+b_2+a_1b_1),\\
(a_1,a_2)\cdot(b_1,b_2)&=(a_1b_1,a_1^2b_2+b_1^2a_2).\\
\end{aligned}
$$
Here $a_1,a_2,b_1,b_2$ are arbitrary elements of $A$, and the operations (on the r.h.s.) involving them are those of $A$.

As $(0,0)$ is the neutral element of addition, we solve from the addition formula that
$$
-(a_1,a_2)=(a_1,a_2+a_1^2).
$$
Therefore the difference
$$
\begin{aligned}
W_2(F)(x_1,x_2)-(x_1,x_2)&=(x_1^2,x_2^2)-(x_1,x_2)\\
&=(x_1^2,x_2^2)+(x_1,x_2+x_1^2)\\
&=(x_1^2+x_1,x_2^2+x_2+x_1^3+x_1^2).
\end{aligned}
$$
Our Witt vector equation $(*)$ is thus equivalent to the system
$$
\begin{cases}
x_1^2+x_1&=f_1,\\x_2^2+x_2+x_1^3+x_1^2&=f_2.
\end{cases}\qquad(**)
$$
The elements of the Galois group $Gal(k(x_1,x_2)/k)$ are then given by adding a Witt vector from $W_2(\Bbb{F}_2)$ to $(x_1,x_2)$. The isomorphism from $\Bbb{Z}_4$ to $W_2(\Bbb{F}_2)$ goes like $0\mapsto (0,0)$, $1\mapsto (1,0)$,
$2\mapsto (1,0)+(1,0)=(0,1)$, $3\mapsto (1,0)+(0,1)=(1,1)$. The corresponding elements of the Galois group are
$$
\begin{aligned}
\tau_0:\,&(x_1,x_2)\mapsto (x_1,x_2)+(0,0)=(x_1,x_2),\\
\tau_1:\,&(x_1,x_2)\mapsto (x_1,x_2)+(1,0)=(x_1+1,x_2+x_1),\\
\tau_2:\,&(x_1,x_2)\mapsto (x_1,x_2)+(0,1)=(x_1,x_2+1),\\
\tau_3:\,&(x_1,x_2)\mapsto (x_1,x_2)+(1,1)=(x_1+1,x_2+1+x_1).
\end{aligned}
$$

With all this in place it is easy to verify that for example $\tau_1$ respects the latter equation of $(**)$:
$$
\begin{aligned}
\tau_1(x_2^2+x_2+x_1^3+x_1^2)&=(x_2+x_1)^2+(x_2+x_1)+(x_1+1)^3+(x_1+1)^2\\
&=x_2^2+x_1^2+x_2+x_1+(x_1^3+x_1^2+x_1+1)+(x_1^2+1)\\
&=x_2^2+x_2+x_1^3+x_1^2.
\end{aligned}
$$

If you feel like it you can verify that $\tau_i\circ\tau_j=\tau_{i+j\bmod4}$ for all the pairs $i,j$:-)

Anyway, from $(**)$ it is obvious that both $k(x_1)/k$ and $k(x_1,x_2)/k(x_1)$ are Artin-Schreier extensions (cyclic of degree two). Furthermore, the automorphism $\tau_2$ has fixed field $k(x_1)$ and generates the Galois group $Gal(k(x_1,x_2)/k(x_1))$.

With all this in place we can then easily calculate the minimal polynomial $m(T)\in k[T]$ of $x_2$ over $k$. The conjugates can be read from the above table, so we just calculate
$$
\begin{aligned}
m(T)&=(T+x_2)(T+x_2+1)(T+x_2+x_1)(T+x_2+x_1+1)\\
&=(T^2+T+x_2^2+x_2)(T^2+T+x_2^2+x_2+x_1^2+x_1)\\
&=(T^2+T+(x_1^3+x_1^2+f_2))(T^2+T+(x_1^3+x_1^2+f_2+f_1))\\
&=T^4+T^2+(T^2+T)f_1+(x_1f_1+f_2)^2+f_1(x_1f_1+f_2)\\
&=T^4+T^2+(T^2+T)f_1+f_1^3+f_1^2+f_2
\end{aligned}
$$
barring a mistake or two. I'm not entirely sure this is what you were looking for. Anyway, the use of Witt vector equations makes it quite clear that the Galois group is cyclic of order $p^2$. All by analogy with the Artin-Schreier case.

Hoping this helps.