The square cover number and the number of horizontal sides I am looking for a geometric upper bound on the square covering number of a rectilinear polygon. 
A square covering of a given polygon is a collection of squares, possibly overlapping, whose union equals the polygon. 
The square covering number of a polygon is the smallest number of squares in a square covering.
To explain the type of bound I am looking for, consider a "polygon" with sides in two directions:

This polygon has only western and southern sides, 4 sides in each direction. It is considered unbounded to the east and north, i.e. the squares in the covering may overflow to the east and north. Its square cover number is also 4:

It is obvious that the square cover number of every such polygon with lines in two directions is equal to the number of sides in one of the directions, e.g. number of horizontal sides, since each horizontal side can be covered by a single square.
Now consider this polyline, which has sides in three directions: west, south and east (it is considered unbounded to the north):

Its square cover number is 4, which is again the number of horizontal sides, since the polygon can be covered by putting a square in each horizontal side from the lowest to the highest:

This time, however, the equality does not hold in all cases: there are polylines whose cover number is smaller (for example, if some of the horizontal sides happen to have the same $y$ value), and there are polylines whose cover number is larger, for example, if there are deep holes like this:

When is the cover number less than or equal to the number of horizontal sides? The following condition seems sufficient to guarantee this: For every pair of vertical sides, their common visible $y$ length is less than or equal to their $x$ distance. This condition prevents deep holes and guarantees that the polyline can be covered by at most one square per horizontal side.
Now consider the general case of a rectilinear polygon with sides in all four directions. The previous sufficient condition does not seem to work. For example, in the following polygon with $k$ teeth, the common visible length of every pair of sides, both horizontal and vertical, is less than or equal to the distance between them, yet the number of horizontal sides is $2k$ and the cover number is $3k-1$ (in the picture $k=5$):

On the contrary, in this similar image:

with the same number of horizontal sides (10), the cover number is only 6.
MY QUESTION: What would be a sufficient condition for having the square cover number bounded by the number of horizontal sides?
Is there a simple geometric condition which allows us to make sure that the cover number of a hole-free rectilinear polygon is at most the number of horizontal sides? 
 A: A sufficient condition could be the following for the case of a polyline bounded on all 4 sides:  
For every pair of vertical sides with common length $y$, let $y_1, y_2,..., y_n$ be the lengths of $n$ sides parallel to the original pair. Then the square cover number is equal to the number of horizontal sides if the horizontal distance between the $y$ lengths, $x$, is equal to $y - \max{\{y_1,y_2,...,y_n\}}$.  
Let me explain. Take for example the rectilinear saw tooth example given. Here is the diagram with the sides labelled (each $y_n$ refers to the segment to its immediate right):

$y_1,...,y_n$ are all equal in this case, but they don't have to be. Here, obviously $x > y - \max{\{y_1,y_2,...,y_n\}}$, so it doesn't work and you must cover $x$ with several smaller squares of length $y_n$.
Now, take this case:

Here, the condition is satisfied for every pair of sides. For the ones labelled with length $y$, the space can be covered with one big square and everything is fine since $x = y - \max{\{y_1,y_2,...,y_n\}}.$ As for each of the teeth, there are no parallel lines in between them and so $y_n$ is nonexistent, meaning they have length 0. The condition then reduces to the equality $y = x$.  
This seems to leave startlingly few rectilinear polygons with these requirements. In fact, the only types seem to be the saw tooth kinds, squares, and kinds of 'embedded' saw tooth figures, like this one:

(Note that in my drawing one of the main teeth is slightly smaller than the other by accident, and so the figure is a bit off.)  
As for the intuitive geometric meaning behind this, It means that not only should the figure have no deep holes (hence no greater total $y$ than $x$), as in the three side polyline, but it should also not have a horizontal hole, where $x$ is greater than $y$ as in the last example in your problem. That leaves equality, which is in the case of a square. But, if the figure sort of 'doubles back' on itself, forming a tooth, then the $x$ length must equal the $y$ length minus this doubling back on itself, called $y_1$. Since each tooth need not be equal to the other in length, then the length subtracted must be the maximum of the two. In fact, if the figure doubles back on itself $n$ times forming $n$ (possibly irregular) teeth, then the condition above results: $x = y - \max{\{y_1,y_2,...,y_n\}}$.
Note that when there are embedded teeth, as in the last example, each "teeth system" must be considered separately.  
It's an interesting problem. I hope this helps.
