Need help understanding example of semidirect product I have come across the following exposition of groups of order $p^3$ via semidirect products:

Let $ p $ be an odd prime. Let $ H = Z_p \times Z_p $ and $ K=Z_p $. We know that the automorphism group $ \text{Aut}(H) $ is isomorphic to a general linear group, $ \text{Aut}(H) \cong \text{GL}_2(F_p) $ and its size is $ |\text{GL}_2(F_p)| = (p^2-1)(p^2-p) $ thus $ p | \text{Aut}(H) $ and by Cauchy's theorem there is an element of $ \text{Aut}(H) $ of order p and hence a nontrivial group homomorphism $ \phi: K \to \text{Aut}(H) $ and so the associated semidirect product $ H \rtimes_{\phi} K $ is a non Abelian group of order $ p^3 $. More specifically, if $ H = \langle a \rangle \times \langle b \rangle $ and $ K = \langle x \rangle $ then $ x $ acts on $ a $ and $ b $ by:
$$ x \cdot a = ab $$ $$ x \cdot b = b $$

With respect to the $ F_p $ basis $ a,b $ of the 2-dimensional vector space $ H $ the action of $ x $ in addititve terms is the linear transform

$$ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} $$

Here is where I am confused: how did they determine the action of $ x $ on $ a,b $? And how did they move from multiplicative notation to additive notation? How to view the group $ H $ as a vector space with basis $ a,b $?
This example confuses me and I would appreciate a more detailed explanation on the transitions they have made. Thanks to all helpers.
 A: Up to a choice of basis, that is essentially the only automorphism of order $p$ of $C_p\times C_p$.
The cyclic group of order $p$ is isomorphic to the integers modulo $p$ under addition; but this is in fact also a field. Whenever you have a field, you can form vector spaces by taking “tuples of elements of the field”, and doing coordinatewise addition and scalar multiplication.  So $H$ “is” a vector space of dimension $2$ over the field of $p$ elements. The basis vectors are $(1,0)$ and $(0,1)$; we identify $(1,0)$ with the generator $a$ of the first copy of $C_p$, and the vector $(0,1)$ with the generator $b$ of the second copy of $C_p$. But if we think of $C_p$ multiplicatively, then $a^k$ corresponds to the vector $(k,0)$ (modulo $p$), $b^k$ to the vector $(0,k)$, and $a^ib^j$ to the vector $(i,j)$. 
Note that $x$ is an automorphism of order $p$ of the vector space $\mathbb{F}_p\times\mathbb{F}_p$ of dimension $2$. Because it is of order $p$, it satisfies the polynomial $t^p-1 = (t-1)^p$ (since we are working in characteristic $p$). So the minimal polynomial divides $(t-1)^p$ and therefore splits. Hence the characteristic polynomial splits and so $x$, viewed as a linear transformation on the space, has a Jordan canonical form. The only two Jordan forms possible (since the only eigenvalue is $1$) are the identity and the matrix given. Since the action of $x$ is not trivial, it does not act like the identity, and so it must be, up to a choice of basis, given by the matrix you display.
This matrix takes the basis vector $(0,1)$ (which corresponds to $b$) and maps it to itself; and takes the vector $(1,0)$ (which corresponds to $a$) and maps it to $(1,1)$, which is the vector sum of $a$ and $b$; but since we are writing the base group multiplicatively rather than additively, that means that $a$ is mapped to $ab$. Thus, the action of $x$ is: $a\mapsto ab$ and $b\mapsto b$. Hence, $x\cdot a= ab$ and $x\cdot b = b$, with $\cdot$ representing the action. 
The move from additive to multiplicative notation and back is a convenience; $H$ is written multiplicatively, but when we think of it as a vector space, we want the underlying operation to be “addition”.
