Why cis-trans isomers exist in certain molecules?

#"Cis-trans"# isomerism can arise in carbon chemistry from the presence of (i) an olefinic bond, #C=C#, or (ii) the presence of a ring structure. This sort of isomerism can also arise in inorganic chemistry also, but I will confine the discussion to organic examples.

Take an olefin with an internal double bond, say 2-butene, #H_3C-CH=CH-CH_3#. Consider the disposition, the geometry, of the methyl groups attached to the vinyl carbons. Because both vinyl carbons have definite geometry, and these carbons CANNOT rotate with respect to each other, 2 geometric isomers are possible; the connectivity is the same, however, the orientation of substituents is different.

With respect to 2-butene, the 2 geometries are, (i) cis -2-butene, with methyl groups on the same side of the double bond; or (ii) trans-2-butene, with methyl groups on different sides of the double bond. This is usually trivial to represent on paper (which of course you will have to do). Please note that the cis and trans isomers, which have both the same formula, AND the same connectivity (i.e #C1# connects to #C2# connects to #C3# etc.), ARE DIFFERENT MOLECULES, with distinct physical and chemical properties.

Likewise, take a disubstituted ring structure, e.g. #"1,2-dimethylcyclohexane"#, #1,2-(H_3C)_2C_6H_10#. The methyl groups can be on the same face of the ring (i.e #"cis"#) or on opposite faces of the ring (i.e. #"trans"#). Because the connectivity is THE SAME in each molecule, these are again geometric isomers, with distinct physical and chemical properties.

How do you rationalize and remember all this? The best way is to make a model, and you are generally allowed to bring model sets into exams. Of course, you have to be able to represent your three-dimensional models on two-dimensional paper. This is a non-trivial exercise, and I would refer to your organic texts, and your lecture notes.