I have problem understanding the direction of the shear stress Is it in the direction of the velocity ? 
If it is, can anybody describe it somehow that matches my physical feeling? 
How the shear stress affects another layer of fluid ? 
I faced a question today that fluid is moving alongside a horizontal tube and and if we write momentum equilibrium we must omit gravity force because it is not in the direction of the velocity, but what about shear stress ? 
 A: Since you are talking about layers of liquid, i am going to assume you are talking about incompressible, laminar pipe flow (i.e. $\mathrm{Re}\ll2000$).

If it is, can anybody describe it somehow that matches my physical feeling?

Shear stress is simply friction between layers of fluid. Imagine rubbing your flat hand across each other, it should be relatively easy to understand that the friction force is in the direction of the movement. This is the same with shear stress. It maybe more convincing to understand that shear stress has units of pressure which can be understood as the friction force exerted per $m^2$ of fluid layer.

How the shear stress affects another layer of fluid ?

Physically, shear stress is the diffusive transport of momentum. As is well-known, diffusion occurs from a high to low concentrations of mass/energy/momentum. More accurately the diffusion transport $j_{\alpha}$ occurs in the direction of a negative mass/energy/momentum concentration gradient:
$$j_{\alpha,m}=-D\partial_{\alpha}\rho \quad j_{\alpha,T}=-k\partial_{\alpha}T \quad j_{\alpha,v_{\beta}}=-\mu\partial_{\alpha}v_{\beta}$$
I write mass/energy/momentum because their diffusive transport is physically similar and there is a clear analogy. The quantity $j_{\alpha,v_{\beta}}$ is what we call the stress; it can be shearing ($j_{y,v_x}$ or $j_{x,v_y}$) but also something else, e.g. ($j_{x,v_x}$ or $j_{y,v_y}$). This something else is a normal (as opposed to shear) contribution to the stress (much like a pressure) which may be important in turbulent or compressible flows. Since we are dealing with incompressible flow you can neglect it.
Now, imagine the pipe and that we have an initially uniform flow; all layers are moving at the same uniform velocity so there is no friction between layers; another way to see this is the gradients are zero so there is no diffusion of momentum between layers. If we now zoom in on the wall, assuming a no-slip condition where the velocity is zero at the wall, we see that there is obviously friction between the wall and the next nearest layer. The next nearest layer will be slowed down by this because it has lost some of its momentum and this has caused friction with the layer above it because that is still going at the uniform speed. This continuous until it reaches the layer exactly at the center of the pipe which sees a layer above and below it with exactly the same (slightly lower) velocity (also known as a symmetry condition). At this point the diffusion has reached an equilibrium and the flow has reached a steady-state. From the perspective of diffusion, friction at the wall is a momentum sink (i.e. source of low momentum concentration) and frictionless layers at the center of the pipe are a momentum source (i.e. source of high momentum concentration).  
Now let's bring it all together.

Is it in the direction of the velocity ?

If you read the pipe flow story carefully you should realize that, while the flow is in the $x$-direction, the diffusive transport of momentum is from the center of the pipe to the wall, i.e. in the $y$-direction. Therefore, the shear stress that is relevant in your equations is:
$$j_{y,v_x} = -\mu\partial_y{v_x}$$
which exerts a friction force in the $x$-direction so belongs in the $x$-component of the force balance, but varies in the $y$-direction.
A: It should help to go back to a definition here. Consider the diagram below:

A material is enclosed between two parallel plates with surface area $A$ and distance $h$ apart from each other.
Assume that a force $F$ is applied to the top plate which now starts moving at constant speed $v$ (the bottom plate is kept stationary).
The equation of motion for this situation is:
$$F=\mu A\frac{\partial v}{\partial h}.$$
We can also re-write this as:
$$\sigma = \mu \gamma.$$
Where $\sigma$ is the shear stress, $\gamma$ the shear rate and $\mu$ the dynamic viscosity.
I hope this helps.
