If crossover didn’t occur and the paired chromatids remained identical, then the four possible combinations would have been as follows:
Reference URL:
https://ib.bioninja.com.au/standard-level/topic-3-genetics/33-meiosis/random-assortment.html
However, the crossover has now produced four non-identical chromatids for one of the pairs of chromosomes, doubling the possible combinations. Therefore, in the above diagram, instead of four identical pairs in each combination of the bottom row, imagine replacing one of each combination with a longer chromatid that has genetic materials of mixed parental origin (eg. In combination 1, instead of two pairs of ‘blue‘ chromosomes, there would be one pair of blue chromosomes and one pair with the longer chromosome having red tips - ie. substitute one of the recombinant chromosomes in the diagram below for its paired nonrecombinant chromosome in the diagram above to get the eight different combinations in the bottom row).
Reference URL:
https://bodell.mtchs.org/OnlineBio/BIOCD/text/chapter9/concept9.6.html
NB: The only difference between combining these diagrams and hand drawing the chromosomes in the original question is that there were two crossover points instead of one, so for the specific answer to your question, imagine the long blue chromatid as having only one red tip instead of two and the long red chromatid as similarly having only one blue tip.