Problem statement
Given
\begin{gather} u_{t}=u_{xx},\\u(t,0)=u(t,2\pi),\\u_{t}(t,0)=u_{t}(t,2\pi),\\u(0,x)=x \end{gather}
find $u$.
Attempt at solution
$$u(t,x)=T(t)X(x)$$ therefore, $$\frac{T'}{T}=\frac{X''}{X}=\lambda.$$
We see according the 2nd and 3rd condition that the function we look for must be periodic. So it cannot be of the form $X=ce^{kx}$ or $X=kx+B$ whereas k isn't $0$. if $y=B$, the 2nd and 3rd conditions apply, and then: $u_{xx}=0$. which is still correct, therefore: $B=X$. which leads to that there is no a linear solution, as no fixed solution equals $x$. So we are left with $cos$, $sin$ solutions. For any $\phi_n$, 2nd and 3rd conditions apply. So we shall now find $T$. $T$ must be in the form $T=ce^{kt}$. Now let's search solution for $X$, it can't be based upon $cos$ function as they are even, therefore we're left with $sin$ functions : $$\int cos(nx)xdx=\frac{(-2nπcos(2nπ)+sin(2nπ))}{n^{2}}=-\frac{2π}{n}.$$ Therefore ,$$X=-\sum_{n\notin\mathbb{Z}}\frac{2\pi\sin(nx)}{n}.$$ And thus, $$u(x,t)=Ce^{kt}\left(\sum_{n\notin\mathbb{Z}}\frac{2\pi\sin(nx)}{n}\right).$$
It is interesting that your initial condition does not satisfy the endpoint conditions. But, that's okay. To have the desired periodicity in $x$, you can use $$ u(t,x) = \sum_{n=-\infty}^{\infty}a_n(t)e^{inx}. $$ This solution satisfies the requirements of periodicity. In order for this to be a full solution, you'll need $u_t=u_{xx}$, which leads to $$ a_n'(t) = -n^2a_n(t) \implies a_n(t) = C_ne^{-n^2 t}. $$ The constants $C_n$ are determined by the initial condition $u(0,x)=x$: $$ x = \sum_{n=-\infty}^{\infty}a_n(0)e^{inx}=\sum_{n=-\infty}^{\infty}C_n e^{inx}. \\ C_n = \frac{1}{2\pi}\int_{0}^{2\pi}xe^{-inx}dx. $$