Do large objects take more time to occupy the same portion of the viewing field when approached, than smaller objects do? First, in more broad terms - I would like to find out if larger objects from longer distances away take more time to appear over the same portion of the viewing field (sight), than smaller objects do from shorter distances away, when approaching them with the same speed.
Example:
I am walking on a straight path. At a short distance in front of me, there is a house measuring 5 m in height. I hold up a coin in front of my face in order to cover it, in other words - block it from my viewing field. The bottom and top edges of the house fit exactly within the diameter of the coin. I approach the house and it's edges start appearing over the edge of the coin. I stop when the height of the house appears double that of the coin (when the edges of the house each extend half of the diameter of the coin from behind it). From the moment the house fits exactly behind the coin, It takes me X amount of time of approaching the house, for it to appear twice as big as the coin in my viewing field.
I move past the house and there is a mountain in front of me, which measures 5,000 m in height. I hold up the same coin in front of my face and at same distance from my face as when I was covering the house. The bottom and top of the mountain fit exactly within the diameter of the coin and thus the mountain is blocked from my sight. I approach it with the same speed with which I approached the house, until the top and bottom of the mountain start appearing over the edge of the coin. I stop when the height of the mountain appears double that of the coin (when the bottom and top of the mountain each extend half of the diameter of the coin from behind it.) From the moment the mountain fits exactly behind the coin, It takes me Y amount of time to approaching the mountain, for it to appear twice as big as the coin in my viewing field.
Question:
Is X greater than Y?
Is the amount of time of approaching which took the mountain to appear twice as big as the coin, from the moment it fit exactly behind the coin - different than - the time it took the house to appear twice as big as the coin, from the moment it fit exactly behind the coin?
What determines the difference?
(my rank doesn't yet allow me to post pictures)
 A: $Y$ is definitely greater than $X$. In fact, $Y/X$ is pretty nearly proportional to the height of the mountain divided by the height of the house, hence, a large number. 
If we imagine the ground as a horizontal line, and the house as a vertical line, and we make you an ant, so that your eye is at ground level, then similar triangles tells you this. 
Relevant reference: there's a (marine) navigation technique called "doubling the angle on the bow" that uses more or less the idea you've described.  
The following picture gives a side view of the ant's-eye view of the world: the short vertical segment at the left is the "coin", which happens to exactly obscure the house (thick blue line) AND the mountain (thick brownish line). When the ant's moved to the position indicated in red, the coin obscures half the house...and the ant happens to have moved halfway to the house. Similarly, when the ant's moved to the green position, the mountain is half-obscured...and the ant has moved halfway to the mountain, which is a LOT farther than halfway to the house. 
This all uses the ant's-eye assumption, but essentially the same argument holds for the case where the eye is 5 or 6 feet above the ground plane --- it's just a little bit messier. 
For your final question -- "What determines the difference?" -- I guess the answer is "geometry" or "the ratio of the mountain height to the house height", depending on how general you meant the question to be. (Another thing that makes all this work is that light travels in straight lines, but you probably knew that, at least implicitly.) 

