And this my basic attempt. My question is how to deal with source in terms of the solution itself? Usually, I solved this source term first but in this case, I don't know how to continue writing...And to confirm if I am on the right way?
Thanks in advance
The PDE is linear homogeneous in $u(x,t)$ the boundary conditions are homogeneous Dirichlet.
I substitute the ansatz $u(x,t)=X(x)T(t)$ into the pde and divide by $u(x,t)$ and obtain
\begin{align} T'(t)/T(t)-k X''(x)/X(x)=\gamma \end{align}
for the above to hold we must have $X''(x)/X(x)=\lambda$, $X(x)$ must satisfy the homogeneous boundary conditions, so we get a family of solutions to the eigenvalue problem \begin{align} X_{n}(x)=\sin\left(\frac{n\pi x}{L}\right),\,\,\,\lambda_{n}=-n^{2}\pi^{2}/L^{2} \end{align}
substituting back we have
\begin{align} \frac{T'_{n}(t)}{T_{n}(t)}-k\lambda_{n}=\gamma \end{align}
which gives $T_{n}(t)=\exp(\left(\gamma+k\lambda_{n}\right)t)$
\begin{align} u(x,t)=\exp(\gamma\, t)\sum_{n=1}^{\infty}c_{n}\exp(k\lambda_{n}t)\sin(\frac{n\pi x}{L}) \end{align}
With hindsight one can also make the change of variables $u(x,t)=\exp(-\gamma t)w(x,t)$, one sees, for which one has
\begin{align} u_{t}=-\gamma\exp(-\gamma t)w(x,t)+\exp(-\gamma t)w_{t}=-\gamma u+\exp(-\gamma t)w_{t} \end{align}
\begin{align} u_{t}+\gamma u=\exp(-\gamma t)w_{t} \end{align}
The PDE in $w(x,t)$ is the one that you are familiar with
\begin{align} w_{t}-kw_{xx}=0 \end{align}
I hope one of these two answers helps you.