Why is this not in 2NF and 3NF form?
If I understand you correctly,

{A} determines {C,F}, and {C}determines {D}, so {A} determines {C, D, F}

Also {B} determines {E},

Put together, this means {A, B} uniquely determines the entire row {A, B, C, D, E, F}. Therefore, columns {A, B} is a key of relation R.

Now, 2NF is violated if a non-key field is determined (functionally dependent) by a proper subset of the key, which is obviously the case here for {B} -&gt; {E}. i.e. part of the key {A, B} alone is sufficient to determine a non-key field {E}. This causes all the redundancy issues etc. and violates 2NF. So R is not 2NF. 

To normalize R to 2NF, we need to decompose R into 

R1 = {A, C, D, F} 


and 

R2 = {B, E}

R3 = {A, B}


Now, {A} is key for R1 and {B} is key for R2, and they have no proper subsets (non-empty). {A, B} is key for R3 and there is no non-key field in R3. Therefore, R1, R2, R3 are in 2NF.

3NF is violated if a non-key field functionally depends on another non-key field. This is the case for R1 where we have {A} -&gt; {C} and {C} -&gt; {D}, which causes similar problems with the violation of 2NF. To fix/normalize this, we can split R1 = {A, C, D, F} further into:

R11 = {A, C, F}


and 

R12 = {C, D}


Now, it's easy to verify that R11, R12, R2, R3 together are in 3NF.