This website may help. Hopefully it is at the appropriate level. If not you will have to be a bit more specific.
When we look at a skeletal representation, we know that each bend of the chain represents #CH_2#, a methylene unit. Where there are 3 #C-C# bonds depicted, a #C-H# or methine bond is specified. If there is an unsaturated bond, i.e. #C=C#, or #C=O#, or #C=N#, OR a ring junction, the unsaturation (the double bond! or ring) will be so represented, and the structure will bear 2 hydrogens less than the saturated structure (this is called 1 degree of unsaturation). An alkyne will bear 4 hydrogens less than the saturated structure; i.e. acetylene, #H-C-=C-H#, versus ethane, #H_3C-CH_3# (i.e. acetylene has 2 degrees of unsaturation). For a saturated molecule (i.e. single bonds only), each junction of the bond represents a methylene unit, a #CH_2# group. For a linear alkane with #n# carbon atoms, there will be #2n + 2# hydrogen atoms - alkanes are saturated. Of course, a carbon chain will waggle all over the place; we usually draw them so as to maximize molecular symmetry. Why? Because it's simpler that way.
Of course, there are a lot of pitfalls, and nuances, which you can't learn all at once. Good luck.
