Newtonian mechanics on an inclined plane Consider a block of mass $m$ on a completely frictionless, fixed inclined plane inclined at an angle $\theta$ as shown here. Neglect any forces other than weight of the block.

So, as I have drawn here, a force $mgsin\theta$ acts on the block along the inclined surface. 
MY QUESTION:
We know that the block will move down and along the surface i.e part of it's acceleration is towards the right. However, we already know that only the weight $mg$ is acting on the block and that has no horizontal component. So, what causes this part - horizontal reaction that we observe? What is the force I'm missing here? Illustrating my question here..

 A: Why are you neglecting any force on the block other than its weight?
Isn't it the normal reaction of wedge on the block that has a horizontal component, giving it an acceleration in that direction? 
A: Your second digram in your question is $\textbf{wrong}$. There is no force $F=mg\sin \theta\cos \theta$ for an object on an inclined plane. 

However, we already know that only the weight $mg$ is acting on the block and that has no horizontal component

I think you are going wrong with this statement that you made in your question. The $mg$ force on an inclined plane is correct the way you have drawn it, and you can resolving this force in two directions, either parallel or perpendicular to the surface of the plane. And more importantly the weight is not the $\textbf{only}$ force on the block, there is also a normal force. 
So really, the normal force has a horizontal component that is giving it an acceleration along the inclined plane. 
A: Suppose we do the force balances in the horizontal and vertical directions rather than in the directions parallel and perpendicular to the incline.  In this case, the vertical component of the normal force is $N\cos{\theta}$ and the horizontal component is $N\sin{\theta}$.  And, if a is the component of acceleration tangent to the plane, then the vertical component of acceleration is $a\sin{\theta}$ and the horizontal component is $a\cos{\theta}$.  So, the force balances in the vertical and horizontal directions are:
$$mg-N\cos{\theta}=ma\sin{\theta}\tag{vertical}$$
$$N\sin{\theta}=ma\cos{\theta}\tag{horizontal}$$
If we multiply the vertical equation by $\sin{\theta}$ and the horizontal equation by $\cos{\theta}$, and add, we obtain:$$mg\sin{\theta}=ma$$ or$$a=g\sin{\theta}$$If we multiply the horizontal equation by $\sin{\theta}$ and the vertical equation by $\cos{\theta}$ and subtract, we obtain$$N=mg\cos{\theta}$$
From the horizontal equation, it is seen that it is the normal force that is causing the horizontal acceleration.
