The droplets on your screen act as plano-convex lenses.
As depicted in the schematic diagram above, the focal length $f$ is the distance between the lens and it's focal point. $f$ depends on the radius of curvature of the lens: A higher radius of curvature results in a larger $f$.
The radius of curvature decreases for smaller droplets due to surface tension (i.e. the bigger the droplet, the "flatter" it is).
The pixels of your screen have a fixed distance $d$ to the droplets (the thickness of your screen's glass). Now if we compare a droplet for which $d>f$ with a droplet for which $d<f$, we notice that the image has to be flipped (cf. with the diagram).
If, however, $d \approx f$, the light of the single pixels is "spread" almost over the whole lens, resulting in the blurry spots you see in the center of the diamond-shaped images.
The diamond shape consists, if you look closely, of the black horizontal lines connecting to the black vertical lines due to distortion of the image. For a drop with perfect size, $d = f$, the diamond shape sould not be visible, because in this case the blurry spot would cover the whole image.
The above also explains why very small and very large droplets appear transparent: For $d \ll f$, it is almost as if there is no lens. For $d \approx 2f$, the image is just flipped, but not magnified.
Also, the color of the blurry spot is coming from the fact that each pixel consists of three single-coloured subpixels.