Using Laplace Transform you have
$$
s^2 \hat{u} - \hat{u}_{xx} = 0
$$
with boundary conditions
$$
\hat{u}(0,s) = \hat{\phi}(s), \quad \hat{u}_x(\pi,s) = 0
$$
wich leads to the solution
$$
\hat{u}(x,s) = \hat{\phi}(s) \frac{\cosh s(\pi-x)}{\cosh \pi s}
$$This is fantastic modulo inverting the Laplace transform, which I think it can be done, but I'm still not clear on the details.
Another way would be taking $v(x,t) = u(x,t) - \phi(t)$ and using separation of variables. This is the battle horse and it's kind of fool proof, but it involves a lot of work.
The most intuitive and, in my opinion, beautiful way to solve the equation is using the fact that for any parallelogram $ABCD$ in the $xt$-plane bounded by four characteristic lines, the sums of the values of u at opposite vertices are equal, that is
$$
u(A) + u(C) = u(B) + u(D) \tag{1}
$$
If we divide the $xt$-plane in regions delimited by the characteristis as shown in the figure
$\hskip1in$
and take $A = (x,t) \in \mbox{II}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,0) \in \mbox{I}$, $D=(x_D,t_D) \in \mbox{I}$, D'Alambert solution implies that
$$
u(x,t) = 0, \quad (x,t) \in \mbox{I}
$$
and using $(1)$, we have that
$$
u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{II}
$$
For III, no wave can reach the region, hence
$$
u(x,t) = 0, \quad (x,t) \in \mbox{III}
$$
For region IV, take $A = (x,t) \in \mbox{IV}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,t_C) \in \mbox{I}$ and $D = (x_D,t_D) \in \mbox{III}$ and equation $(1)$ implies that
$$
u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{IV}
$$
For Region V we take $A = (x,t) \in \mbox{V}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{III}$ and $D = (x_D,t_D) \in \mbox{III}$ we have that
$$
u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{V}
$$
Region VI is more interesting: taking $A = (x,t) \in \mbox{VI}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,t_C) \in \mbox{II}$ and $D = (\pi,t_D) \in \mbox{VI}$, equation $(1)$ implies that
$$
u(x,t) + u(x_C,t_C) = u(0,t_B) + u(\pi,t_D) \, \Longrightarrow
$$
$$
u(x,t) = \phi(t_B) - \phi(t_C - x_C) + u(\pi, t_D)
$$
but
\begin{align}
t_B &= t - x \\
t_D &= t + x - \pi\\
x_C &= \pi - x\\
t_C &= t - \pi
\end{align}
and then
$$
u(x,t) = \phi(t-x) - \phi(t + x -2\pi) + u(\pi,t + x - \pi)
$$
Now, using $u_x(\pi,t) = 0 = -2 \phi'(t - \pi) + u_t(\pi,t)$
we have
$$
u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{VI}
$$
In region VII, taking $A = (x,t) \in \mbox{VII}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{IV}$ and $D = (\pi,t_D) \in \mbox{VI}$, equation $(1)$ implies that
$$
u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{VII}
$$
So far, it's easy to understand the results of all regions. For I and III, there is no wave, for II, IV and V, there is only the wave originating from the boundary $x = 0$. In VI there are two waves, the one originated at II plus the reflection on the boundary $x = \pi$. In VII, there is the wave from $x=0$ and region VI. This logic tells us that in VIII there will be three waves: the one from the boundary $x = 0$, the one coming from VI and it's reflection. To see this, we take $A = (x,t) \in \mbox{VIII}$, $B = (0,t_B) \in \mbox{VIII}$, $C = (x_C,t_C) \in \mbox{VI}$ and $D = (\pi,t_D) \in \mbox{VI}$,
\begin{align}
u(x,t) &= \phi(t_B) - \phi(t_C - x_C) - \phi(t_C + x_C -2\pi) + 2\phi(t_D - \pi)\\
&= \phi(t-x) + \phi(t + x - 2\pi) - \phi(t - x - 2\pi), \quad (x,t) \in \mbox{VIII}
\end{align}
Why the change of sign on the reflecting wave? one might ask. The answers is simple: the boundary $x = 0$ is hard (Dirichlet), while the boundary in $x = \pi$ is soft (Neumann).
For region IX, we take $A = (x,t) \in \mbox{IX}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{V}$ and $D = (\pi,t_D) \in \mbox{IX}$,
\begin{align}
u(x,t) &= \phi(t_B) - \phi(t_C-x_C) + u(\pi,t_D)\\
&= \phi(t-x) - \phi(t + x - 2\pi) + u(\pi, t + x - \pi)
\end{align}
Again, using the boundary condition in $x = \pi$ we have $u(\pi,t) = 2\phi(t - \pi)$ for $(x,t) \in \mbox{IX}$ and
$$
u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{IX}
$$
There is clearly a pattern arising in the triangular regions. In the parallelograms, a little more work must be performed, but all in all, the table is set to propose a general solution by induction.
Can you finish it off?