Abigail: I think that you should have enough information from the links (vide infra) I provided to get you through most of what bothers you. It takes time to get these issues straight. Structures 1 represent the isopropoxy pyranosides of D-glucose ($\ce{C5}$-(R)). Assuming a chair conformation in 1a, all of the groups alternate, i.e.,up,down,up, etc., around the ring. All of the substituents are equatorial. Structure 1b is the chair conformation of 1a. The isopropoxy group is in the β-configuration, meaning that it is above the plane of the ring as drawn in 1a and 1b. The Fischer projection of this D-glucopyranoside is shown in structure 1c. By convention, the isopropoxy group of the β-configuration is drawn on the left side of the Fischer projection.
As @Mathew has noted, your left hand structure is a pentose. But if the hemiacetal OH were replaced by -$\ce{CH2OH}$ of the same configuration, then structures 2 would be the same as structures 1 except for the configuration at $\ce{C2}$. The $\ce{C2}$ epimer of D-glucose is D-mannose wherein the -OH group at $\ce{C2}$ is axial.
Now your right hand structure has some issues. It is a heptapyranoside as shown in structure 3a given the configurations in your structure with the addition of the configuration at the anomeric carbon based upon what your structure displays. Structure 3b is the chair conformation of 3a. Again, by convention, when the methoxy group is drawn on the right of the Fischer projection, it means that the methoxy group has the α-configuration. In keeping with the Fischer convention, structure 3c is the correct representation with the correct configurations.

Drawing Cyclohexanes:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Conformation
http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/cyclohexane.pdf
Carbohydrates:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#carbohydrates