Figure out the component of a value in X and Y coordinates using trigonometry. Alright. It's been long that I studied trigonometry and did Laws of Motion and Free Body Diagrams, and I was decent good at them, but somehow I am having trouble in understanding the following.
Note : I will try to understand the problem statement as well as I can without the use of figures and appropriate notations since I dont know how to use them. But if it is somehow unclear, please let me know.
I get that in an $XY$ plane, if I have a straight line-segment $y=x$ from $(0,0)$ to $(1,1)$ then it means that the component of that line is $1$ unit along X axis and $1$ unit along Y axis. i.e.
The length of line segment between $(0,0)$ and $(1,1)$ is $\sqrt{2}$. Now
Along X axis, value = Length$\cdot \cos(\theta)$ and with $\theta = 45$, we get value along $X = 1$.
Similarly, we get value along Y = Length$\cdot \sin(\theta) = 1$.
I understand the above concept.
Now, if I extend the same concept to Forces. 

Referring to the above image, the force in the X direction is $mg\sin 30$ and in the Y is $mg\cos 30$. 
As I understand, the above force is $mg \sin(30)$ along X is because of the following
Force = $m \cdot g$. Mass is going to be the same in each direction so take that out.
$g$ is basically, distance (displacement) divided by time square. Now time is going to be the same in each direction so we can take that out too. This leaves us with
Force is proportional to Displacement(Distance). Now this is going to be different in different dimensions, that is why we derive the component along X and Y for distance. Hence the $\cos(30)$ and $\sin(30)$ that we see is because of the distance in the Force.
Firstly, am I correct in my understanding of the above?
Now, let's say our force was defined as.
F = (distance)(distance) instead of the usual MassAcceleration.
In this case, if I were to figure out the Force along X and Y, then my 
F along X = F.sin(30).sin(30) instead of F.sin(30). 
F along Y = F.cos(30).cos(30) instead of F.cos(30). Right?
Is the above understanding correct?
Also, how to know if a particular physical quantity will behave differently in a particular dimension. In my original formula, I did not consider Mass and Time to be affected by dimensions, but that was more out of intuition (or maybe some other reason that I dont know). But distance seemed like a quantity that will change in different dimensions.
Or is it more to do with the fact that Mass and Time dont have any directional concept associated with them?
 A: The reason that we can decompose the force $mg$ into components $mg\sin(30^\circ)$ and $mg\cos(30^\circ)$ is not because the concept of force has distance embedded in it (through the distance/time$^2$ in $g$). Rather it is simply because force is a vector quantity. Vectors arise often in physics, and it is important to make sense of them.
Before we can discuss vectors, we have to discuss coordinate systems. This is a way of defining a set of numbers (coordinates) to the system so that every point can be cataloged, so to speak, according to this system. Many systems (like the one you describe here) use a Cartesian coordiante system which consists of two (or more) axes which are perpendicular to each other, and each point is referenced according to its distance from each of these axes. We usually label the axes $x$ and $y$. We are used to $y$ being vertical and $x$ being horizontal. But this doesn't have to be the case. Depending on the problem you are trying to solve, it may make sense to rotate the axes. For instance, in your problem, we could put the $x$ axis in the plane of the ramp, with the $y$ axis perpendicular to it.
Now vectors. Vectors can be represented by a coordinate in the coordinate system (this is actually not generally true, but it will be in introductory physics, so will stick to that). For instance, a vector might have the representation $A=(3,4)$, which can be visualized as an arrow starting at the point $(0,0)$ and going out to the point $(3,4)$. We can also say that this vector is $A=A_x+A_y$ where $A_x=(3,0)$ and $A_y=(0,4)$. This is helpful because $A_x$ and $A_y$ are in the direction of the coordinate axes. Here, $A_x$ and $A_y$ are called components of $A$. But it is not always so easy to find the components. Suppose I said "$A$ is the vector which has a magnitude of $r=5$ and makes an angle of $\theta=36.9^\circ$ above the horizontal axis. This means that $A$ is represented by an arrow of length $5$ at this particular angle pointing slightly up above the horizontal. To find the components, we use trigonometry. By drawing a triangle like the one in the figure below, we find that the components are $A_x=(r\cos\theta,0)=(4,0)$ and $A_y=(0,r\sin\theta)=(0,3)$.

Note that I never had to assume that $A$ had anything to do with physical length here.
So let's go back to your problem. The force of gravity is a vector, and you want to compare it to the force of friction and acceleration (in the plane of the ramp), and the normal force (perpendicular to the ramp). This is a problem because these forces and acceleration are in line with a coordinate axis, but the force of gravity is not. To fix this, we just use the same trigonometry trick to break gravity up into its components: $F_g=(F\cos\theta,0)+(0,F\sin\theta).$
One final note. You may think it odd that I put a second coordinate which is just zero in each of the components (i.e., I wrote $(F\cos\theta,0)$ instead of just $F\cos\theta)$. This is just me being formal. The great thing about breaking every vector up into its components is that no you can compare all the components in the $x$ direction without worrying about those in the $y$ direction, and vice-versa. This is why components are useful. 
In summary, breaking up a force into different pieces has to do with the fact that it is a vector, and not physical distance being part of the units of force. All vectors can be decomposed this way, regardless of the units associated with them. There are a lot of vectors in physics (and other subjects), and it is always useful to know how to break them up into components.
A: Either using vectors or trigonometry the rational doesn't involve traveling distances. 
Since the angle in blue is a right angle (90), the red and black triangles are similar . 
If you decompose the $F=mg$ into its normal ($F_N$) and tangential($F_T$) components  
\begin{equation}
F_N = F cos(30)\\
F_T = F sin(30)
\end{equation}
As you can see $F_T$ is the projection on the $x$-axis and it's positive.  $F_N$ is the projection on $y$-axis and it's negative.
 
Finally, 
\begin{equation}
\sum F_x = F_T - F_f = 0 \\
\sum F_y = N - F_N = 0 
\end{equation}
A: You can't just decompose force into distancedistance. The fact that force is massacceleration is a very fundamental notion. Furthermore, the units don't even compare, since the unit of force is $N=\frac{kg\cdot m}{s^2}$ and you are proposing a unit of $m^2$.
I think you may be getting caught up on how the arrows are drawn. The factor $\cos(30^\circ)$ is because we are decomposing the direction of the force into a different set of axes. Gravity points towards the earth, which is of course always true. The way you define the axis means that we must use trig to find the component of this force along each axis. If you calculate the resulting force as the hypotenuse of the triangle formed by these two forces, you would of course get $\sqrt{(mg\sin 30^\circ)^2+(mg\cos 30^\circ)^2}=mg$.
I hope that answers your question somewhat. I think you may be overthinking the way these forces work. In summary, the force is always mg, always points down, and we can break that vector into component vector in many different ways. They will always have the original force times a trig function.
