What happens when two bar of different material are joined and their opposite ends are heated to same temperature. How would the heat transfer work? Imagine two bar, one of copper and another of steel of same length and cross section are joined together end to end. Now the bars are insulated except the end such that no other forms of heat transfer occur except conduct. Now the opposite end of the joined bar are both heated to say 1000°C.
Now if one dimensional conduction heat transfer occurs, I imagine it would occur from both side toward the joined section. If a steady state is reached, there should be a point where both the temperature profile meets, in that case where would the heat flux propagated from both sides go in that point?
 A: Consider the Fourier heat equation:
$$T_t=\alpha \nabla^2T$$
or in one dimension only:
$$T_t=\alpha T_{xx}$$
Steady state means temperature no longer time-evolves, that is $T_t=0$, so:
$$\alpha T_{xx}=0$$
and assuming that $\alpha \neq 0$, then:
$$T_{xx}=\frac{\mathbf{d^2}T}{\mathbf{d}x^2}=0\tag{1}$$
Note that something important has happened: $T$ is no longer dependent on the thermal diffusivity $\alpha$. So in our case it doesn't matter if the bar is made of one material, two or even many! It doesn't even matter what kind of material the bar is made of.
ODE $(1)$ can be solved easily.
$$T_{xx}=\frac{\mathbf{d^2}T}{\mathbf{d}x^2}=0 \Rightarrow \frac{\mathbf{d}T}{\mathbf{d}x}=C_1$$
Where $C_1$ is an integration constant.
Now:
$$\Rightarrow \frac{\mathbf{d}T}{\mathbf{d}x}=C_1\Rightarrow T(x)=C_1 x+C_2$$
Where $C_2$ is the second integration constant.
The integration constants are found by applying the boundary conditions, usually the temperatures at the extremes of the bar e.g. $T(0)=T_0$ and $T(L)=T_L$.
If, as in your case:
$$T_0=T_L$$
Then it shows that:
$$T(x)=T_0=T_L$$

If a steady state is reached, there should be a point where both the
  temperature profile meets, in that case where would the heat flux
  propagated from both sides go in that point?

No. Steady state means there is no more net heat flux. Otherwise the point towards which the heat flows would still be heating up, which cannot be described as 'steady state'.
