I didn't get what this function $\phi$ is
It's not a function, just a fixed number.
is and how come the conditional distribution has mean equal to $\phi{x_{t-1}}$
Because, assuming $\epsilon_t$ is a iid $N(0,1)$ (white gaussian noise) :
$E(x_t |x_{t-1} ) = E(\phi x_{t-1} + \epsilon_t |x_{t-1}) = \phi x_{t-1} + E(\epsilon_t) = \phi x_{t-1}$
Further, assuming $x_t$ is weakly stationary, so that $E(x_t^m)=E(x^m_{t-1})$, we have
$$E(x_t) = \phi E(x_{t-1}) + E(\epsilon_t) \Rightarrow E(x_t) (1-\phi) = 0 \Rightarrow E(x_t)=0$$
and squaring both sides:
$$E(x_t^2) = E(\phi x_{t-1}+ \epsilon_t)^2 = \phi^2 E( x^2_{t-1}) + E(\epsilon_t^2) $$
(the cross terms vanishes because $\epsilon_t$ is independent of $x_{t-1}$) and so the variance of $x_t$ is given by
$$E(x_t^2) (1 -\phi^2) = E(\epsilon_t^2) \Rightarrow E(x_t^2) = \frac{1}{1 -\phi^2} $$
Added: Here's a derivation for the joint distribution. Let's take $n=4$ , and $a = \phi$
(less typing).
Then
$$p(x_1 x_2 x_3 x_4 ) = p(x_4 |x_3) p(x_3 |x_2) p(x_2 |x_1) p(x_1)$$
The first factor is $N(a \, x_3, 1)$ and the same for the rest, until the last which is $N(0,1/(1-a^2))$, hence
$$ p(x_1 x_2 x_3 x_4 ) = \frac{1}{\left(\sqrt{2 \pi}\right)^4 }
\exp{\left[ -\frac{1}{2} \left(
(x_4- a x_3)^2
+ (x_3- a x_2)^2
+ (x_2- a x_1)^2
+ (1-a^2) \, x_1^2\right)\right]}
$$
The term inside parentheses can be expanded as
$$x_4^2 - 2 a x_3 x_4 + (1+a^2) x_3 - 2 a x_3 x_2 + (1+a^2) x_2 - 2 a x_2 x_1 + x_1^2$$
or, in matricial form:
$$
\begin{pmatrix}
x_1 & x_2 & x_3 & x_4 \\
\end{pmatrix}
\begin{pmatrix}
1 & -a & 0 & 0 \\
-a & 1 +a^2 & -a & 0 \\
0 & -a & 1 +a^2& -a \\
1 & 0 & -a & 1
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4 \\
\end{pmatrix}
$$
So, finally $$p({\bf x}) = \frac{1}{\left(\sqrt{2 \pi}\right)^4} \exp{\left( -\frac{1}{2} {\bf x^T Q x } \right)}$$
which is the formula a joint 4-D gaussian, with zero mean and covariance $\Sigma = Q^{-1}$ And so we have found the precision matrix (inverse of the correlation matrix).