How exactly does a Max 2 Sat reduce to a 3 Sat?
It seems that the lecture note has abused the symbol $c_i$. You should check the original paper.



Replacing each clause $C_i = (a_i \lor b_i \lor c_i)$ of 3CNF by the collection of 2CNF clauses:

$$(a_i), (b_i), (c_i), (d_i), (\bar{a_i} \lor \bar{b_i}), (\bar{a_i} \lor \bar{c_i}), (\bar{b_i} \lor \bar{c_i}), (a_i \lor \bar{d_i}), (b_i \lor \bar{d_i}), (c_i \lor \bar{d_i})$$

where $d_i$ is a new variable.




  Theorem: If an assignment $\alpha$ satisfies $C_i$, then exactly seven of the ten clauses in the 2-CNF collection can be satisfied.


Proof: $\alpha$ satisfies $C_i$, then either one, two, or three of $a_i, b_i, c_i$ must be true. In all cases, there is a truth setting for $d_i$ causing exactly seven of the clauses to be satisfied. For example, suppose $a_i = \top, b_i = c_i = \bot$, then $d_i = \bot$ causes 7 clauses to be satisfied.


  Theorem: If an assignment $\alpha$ does not satisfy $C_i$, then exactly six of the ten clauses can be satisfied. 


Proof: $\alpha$ does not satisfy $C_i$, then $a_i = b_i = c_i = \bot$. In this case, if $d_i = \top$, 4 clauses are satisfied; if $d_i = \bot$, then 6 clauses are satisfied.