You know that $\varphi$ is a homomorphism, and you know that $\varphi(1)=4$. Therefore
$\begin{align*} \varphi(2)&=\varphi(1+1)=\varphi(1)+_7\varphi(1)=4+_74=1\\ \varphi(3)&=\varphi(2+1)=\varphi(2)+_7\varphi(1)=1+_74=5\\ \varphi(4)&=\varphi(3+1)=\varphi(3)+_7\varphi(1)=5+_74=2\\ \varphi(5)&=\varphi(4+1)=\varphi(4)+_7\varphi(1)=2+_74=6\\ \varphi(6)&=\varphi(5+1)=\varphi(5)+_7\varphi(1)=6+_74=3\\ \varphi(7)&=\varphi(6+1)=\varphi(6)+_7\varphi(1)=3+_74=0\;,\text{ and}\\ \varphi(8)&=\varphi(7+1)=\varphi(7)+_7\varphi(1)=0+_74=4\;. \end{align*}$
Here I’m using the homomorphism property: $\varphi(m+n)=\varphi(m)+_7\varphi(n)$ for all $m,n\in\Bbb Z$.
Clearly the values of $\varphi(n)$ will cycle through the pattern $4,1,5,2,6,3,0$ repeatedly. The length of the cycle is $7$, so every seventh value of $\varphi(n)$ will be $0$, starting with $\varphi(7)$; from that it’s not hard to see that
$\ker\varphi=\{n\in\Bbb Z:\varphi(n)=0\}=\{n\in\Bbb Z:7\mid n\}=7\Bbb Z\;,$
the set of multiples of $7$.
To say that $4$ has order $7$ in $\Bbb Z_7$ just means that the smallest positive integer $n$ such that $\underbrace{4+_7\ldots+_74}_n=0\text{ in }Z_7$ is $n=7$. We saw this in the chart above: starting with $4$ and repeatedly adding $4$ produced in turn $4,1,5,2,6,3,0$, the cycle that we already noted, and it wasn’t until we’d added together seven $4$’s that we got the additive identity $0$ of $\Bbb Z_7$.
They were using the fact that $\varphi$ is homomorphism when they calculated $\varphi(25)$. First, $25=21+4$, so by the homomorphorphism property $\varphi(25)=\varphi(21)+_7\varphi(4)$. Now $21$ is a multiple of $7$, so $21\in\ker\varphi$, and $\varphi(21)=0$, and therefore $\varphi(25)=0+_7\varphi(4)=\varphi(4)$. Then they split $4$ as $1+1+1+1$ and used the homomorphism property again:
$\begin{align*} \varphi(4)&=\varphi(1)+_7\varphi(1)+_7\varphi(1)+_7\varphi(1)\\ &=4+_74+_74+_74\\ &=(4+_74)+_7(4+_74)\\ &=1+_71\\ &=2\;. \end{align*}$