By separation of variables,
\begin{equation*}
u(x,t)= X(x) T(t)
\end{equation*}
\begin{align*}
& 4X(x) T''(t)= X''(x)T(t), \\[3pt]
& X(0)T(t)=X(2)T(t)=0,
\end{align*}
\begin{align*}
& 4\frac{ T''(t)}{T(t)}= \frac{X''(x)}{X(x)}, \\[3pt]
& X(0)=X(2)=0.
\end{align*}
\begin{equation}
\frac{T''(t)}{T(t)}=-\frac{1}{4}\lambda,\qquad \frac{ X''(x)}{X(x)}=-\lambda
\end{equation}
This is a Dirichlet-Dirichlet boundary eigenvalue problem for $X(x)$, we have
\begin{align*}
& \lambda_n = \frac{\pi^2n^2}{4}&& n=1,2,\ldots, \\[3pt]
&X_n(x)=\sin (\frac{\pi n x}{2}).
\end{align*}
Then plug $\lambda$ into (5) and solve for $T(t)$,
\begin{equation*}
T_n(t)=A_n \cos(\frac{\pi n}{4}t ) + B_n \sin (\frac{\pi n}{4}t )
\end{equation*}
Therefore,
\begin{equation*}
u_n(x,t)= \bigl( A_n \cos(\frac{\pi n}{4}t )+ B_n \sin (\frac{\pi n}{4}t )\bigr) \cdot
\sin (\frac{\pi n x}{2})
\,\,\,n=1,2,\ldots
\qquad\end{equation*}
\begin{equation}
u(x,t)= \sum_{n=1}^\infty \bigl( A_n \cos(\frac{\pi nt}{4} )+
B_n \sin (\frac{\pi nt}{4} )\bigr) \cdot
\sin (\frac{\pi n x}{2}).
\end{equation}
\begin{align}
&\sum_{n=1}^\infty A_n \sin (\frac{\pi n x}{2})=f(x),\\
&\sum_{n=1}^\infty \frac{\pi n}{4} B_n \sin (\frac{\pi n x}{2} )=0.
\end{align}
Calculate the coefficients $A_n$
\begin{multline*}
A_n = \int_0^2 \sin (\frac{\pi n x}{2}) f(x)\, dx = \int_0^1 x\sin (\frac{\pi n x}{2}) \,dx + \int_1^2 (2-x)\sin (\frac{\pi n x}{2}) \,dx = \frac{2\sin \frac{\pi n x}{2}}{\bigr(\frac{\pi n}{2}\bigl)^2} = (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2}, \,\,\,\,0 \,\,\,\text{when n is even}
\end{multline*}
I also get $B_n = 0$,
The solution is given by
\begin{equation}
u(x,t)= \sum_{n=1 \,\text{odd} }^\infty (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2} \cos(\frac{\pi nt}{4})\cdot
\sin (\frac{\pi n x}{2}).
\end{equation}
