Continuous dependence of the solution from the initial data - Meaning I am looking the following: 
What do we mean by the sentence "continuous dependence of the solution of the heat equation from the initial data, on the interval $[0,2\pi]$ with Dirichlet boundary data $0$, for $t>0$ ? Does something equivalent hold for negative times? 
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I have done the following: 
Let $u_1(x,t)$ and $u_2(x,t)$ $C^2$ solutions of the problem ($0\leq x\leq 2\pi, \ t\geq 0$) 
\begin{align*}&u_t=ku_{xx} \\ &u(0,t)=a_1(t) \\ &u(L,t)=b_1(t) \\ &u(x,0)=f_1(x) \end{align*} and 
\begin{align*}&u_t=ku_{xx} \\ &u(0,t)=a_2(t) \\ &u(L,t)=b_2(t) \\ &u(x,0)=f_2(x) \end{align*} where $L=2\pi$. 
If for some $\epsilon>0$ we have $|f_1(x)-f_2(x)|<\epsilon \ \forall x$, $0\leq x\leq L$ and $|a_1(x)-a_2(x)|<\epsilon$ and $|b_1(x)-b_2(x)|<\epsilon$ $ \ \forall t$, $0\leq t\leq T$ then $|u_1(x,t)-u_2(x,T)|<\epsilon \ \forall x,t$ with $0\leq x\leq 2\pi, \ 0\leq t\leq T$. 
That means that little errors at the initial data don't produce big errors at the solution. 
Is everything correct so far? 
 A: You made already a good start.

continuous dependence of the solution of the heat equation from the initial data, on the interval $[0,2\pi]$ with Dirichlet boundary data $0$, for $t>0$ 

The initial data refers only to $f$, and Dirichlet boundary data $0$ means that we only need to consider the case $a(t)=b(t)=0$.
Then, as Jeff said in the comments, continuous dependence of the solution from the initial data
refers to the following:
For all $\epsilon>0$ there exists an $\delta>0$ such that
$$
  |f_1(x)-f_2(x)|<\delta \;\forall x, 0\leq x\leq L
  \quad\Rightarrow\quad
  |u_1(x,t)-u_2(x,T)|<\epsilon \;\forall x,t, 0\leq x\leq 2\pi, \ 0\leq t\leq T 
$$
is true.
The difference to your version is that you had the same value for $\epsilon$ and $\delta$.
However, in the case of the heat equation with $t>0$ it is possible to show
the above statement by choosing $\delta=\epsilon$.
(This would then be very similar to the statement in the second-to-last paragraph in your question).
So, if you are able to show that
$$
  |f_1(x)-f_2(x)|<\epsilon \;\forall x, 0\leq x\leq L
  \quad\Rightarrow\quad
  |u_1(x,t)-u_2(x,T)|<\epsilon \;\forall x,t, 0\leq x\leq 2\pi, \ 0\leq t\leq T 
$$
is true for all $\epsilon>0$, then you have shown that the solution depends continuously on the initial data.

Does something equivalent hold for negative times? 

No, for negative $t$ the solution of the heat equation does not depend continuously on the initial data.
This can be shown with a counterexample.
We consider the case $T=-1$ and that $t\in [T,0]$.
We also set $k=1$.
As the initial data we choose $f_1(x)=1/n \sin(nx)$ for some $n\in\mathbb N$.
Then it can be shown that the solution $u$ is given by
$$
  u_1(x,t) = 1/n e^{-n^2 t} \sin (nx).
$$
We also choose  $f_2(x)=0$ with has the solution $u_2(x,t)=0$.
Then it can be shown that
$$
 \max_{0\leq x\leq L} | f_1(x)-f_2(x) | = 1/n
 \qquad\text{and}\qquad
 \max_{0\leq x\leq L} | u_1(x,T)-f_2(x,T) | = 1/n e^{-n^2 T} = 1/n e^{n^2}
$$
So for big $n$, a small error of $1/n$ in the initial data can cause a very big error at the solution at time $T=-1$.
