Understanding the resulting image matrix when differentiating image
Your interpretation is correct: directional derivation operators highlight variation in a given direction. Here, you use the $2$-point discrete derivative in the $x$-direction (along image rows). It may emphasize vertical features.

First, such operators indeed extend the initial image range. However, one often uses the absolute value of the derivative to compute a gradient. With the modulus, the derivative values fall back in the initial image range. 

Second, as $[1 \;-1] = -[-1\; 1]$, and since convolution is a linear operator, you will get the opposite result as in the previous case. And it won't matter if you take the absolute value.