What does $\left|x,t\right>$ actually mean (Heisenberg picture)? I am pretty much confused with this notation I believe. The Heisenberg states are denoted by $\left|x,t\right>$ and the Schrodinger states are given by $\left|x(t)\right>$. It seems like both of these are parameterised with time, but the Schrodinger state is parameterised through the position vector,
But then in the case of Heisenberg picture, is the state parameterised using two variables $x,t$? What is this Linear Vector space 
Secondly, in the case of position operator in the Heisenberg picture $\hat X_H(t)$ the eigenvalue equation is given like this,
$$ \hat X_H(t)\left|x,t\right> = x\left|x,t\right> .  $$
So does that mean that for every time $t$ of $\hat X_H(t)$ there is a eigenvalue $x$ with the eigenvector $\left|x,t\right>$. So does that mean 
$$ \hat X_H(t')\left|x,t\right> = 0 .$$
 A: I think your confusion stems from the difference between the two pictures. For the Schrodinger picture the states $\left|{x(t)}\right>$ can evolve in time, while the operators are fixed. However for the Heisenberg picture, the states are fixed, and do not evolve with time, but rather the operators evolve with time.
Schrodinger picture: Given some base ket, $\left|x\right>$, we can evolve it in time to get $\left|x(t)\right> = U(t)\left|x\right>$, where $U(t) = e^{-iHt / \hbar}$ is the time evolution operator. The eigenvalue equation is now:
$$ X \left|x(t)\right> = x(t)\left|x(t)\right> $$
Heisenberg picture: Now the state is fixed $\left|x\right>$, but we evolve the operator. $X(t) = U^\dagger(t) X U(t)$, where $U(t)$ is the the same time evolution operator as before. The eigenvalue equation now reads:
$$ X(t) \left|x\right> = x(t)\left|x\right> $$
EDIT:
So to answer the first two questions, the state is labeled as $\left|x,t\right>$ to indicate it is a fixed state at some point in time $t$, it is still part of the Hilbert space. For the next two questions, for every time $t$ of $X(t)$, there is an eigenvalue $x(t)$. I don't know what time you are referring to by $t'$, but in general no, is is not zero.
A: I) Recall that in the Heisenberg picture, operators [such as e.g. the position operator $\hat{X}(t)$] evolve in time $t$, while states (kets & bras) are independent of time $t$.
In particular, an instantaneous position eigenstate $| x_0,t_0 \rangle_H $ in the Heisenberg picture does not depend of time $t$, cf. Ref. 1. An instantaneous position eigenstate satisfy
$$\begin{align} \hat{X}(t\!=\!t_0)| x_0,t_0 \rangle_H~=~&x_0| x_0,t_0 \rangle_H, \cr {}_H\langle x_1,t_0 | x_2,t_0 \rangle_H~=~&\delta(x_1\!-\!x_2).\end{align}\tag{1} $$
It is important to stress that there are no requirements to $| x_0,t_0 \rangle_H$ for $t\neq t_0$.
II) One typical application of instantaneous position eigenstates [e.g. in connection with the timeslice procedure for the Feynman path integral, cf. Ref. 1] is to decompose the unit operator ${\bf 1}$ via an integral representation of instantaneous position eigenstates
$$ {\bf 1}~=~ \int_{\mathbb{R}} \! d x_0 ~|x_0,t_0 \rangle_{H~H}\langle x_0,t_0 |.\tag{2} $$
References:

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*J.J. Sakurai, Modern Quantum Mechanics, 1994; Section 2.1 and Section 2.5.

