Game Theory (Prisoner's Dilemma): Why dominant strategy yields lower payoff than other possible actions? From class notes:

Dominated Strategy
Action $i$ is a dominated strategy for player $j$ if it yields a lower
  payoff than at least one other actions available to player $j$ for
  every possible actions by all other players. So action $a^{j}_{i}$ is
  a dominated strategy for player $j$ if:
$V^{j}(a^{1}_{j},...,a^{j}_{i},..., a^{M}_{l}) < V^{j}(a^{1}_{j},...,a^{j}_{k},..., a^{M}_{l})$
for at least one $k \neq i$ and all $l$


In the image above, the dominated strategy for both player is $({W, W})$. I understand that the payoff value for picking W is $(6, 6)$ for both players as the other highest payoff would be to pick $NW$ which gives a $(7,3)$ for player 1 or $(3, 7)$ for player 2.
But I don't understand the dominant strategy.

Dominant Strategy
{...}
$V^{j}(a^{1}_{j},...,a^{j}_{i},..., a^{M}_{l}) >
V^{j}(a^{1}_{j},...,a^{j}_{k},..., a^{M}_{l})$

From the same image above, now playing $(NW, NW)$ is the dominant strategy for both players. Given that picking $NW$ yields $(4, 4)$. The payoff value is much lower than other available actions. Yet, the dominant strategy states that:

action $i$ is a dominant strategy  for player $j$ if it yields  a
  higher payoff than  any other actions available to  player $j$ for
  every possible actions by all other  players.

 A: I now understand the reason why the dominated strategy is $W$ for both players.
If player 2 chooses $W$, then the payoff value would be $6+3=9$. 
Comparably, choosing $NW$ would yield $7+4=11$.
Thus, $V^{2}(W) < V^{2}(NW)$. This is the definition of a dominated strategy.
Vice versa, $V^{2}(NW) > V^{2}(W)$ would be the dominant strategy, picking $NW$.
A: By definition,

action $i$ is a dominant strategy  for player $j$ if it yields  a
  higher payoff than  any other actions available to  player $j$ for
  every possible actions by all other  players.

There are multiple quantifiers there; it might be clearer this way:

Action $i$ is a dominant strategy  for player $j$ if
  for every possible selection of actions by all other players,
  action $i$ it yields  a higher payoff to player $j$
  than  any other action available to  player $j$.

So consider action $NW$ for player $1.$
I'll write $NW_1$ for that action to distinguish it from the action $NW_2$ that can be taken by player $2.$
There is only one other action available to player $1,$ namely $W_1,$
so $NW_1$ is dominant if it is true that no matter which action player $2$ takes,
$NW_1$ yields a higher payoff to player $1$ than $W_1$ does.
If player $2$ takes action $W_2$, then $W_1$ pays $6$ to player $1$ and $NW_1$ pays $7$ to player $1.$ So $NW_1$ yields the higher payoff.
If player $2$ takes action $NW_2$, then $W_1$ pays $3$ to player $1$ and $NW_1$ pays $4$ to player $1.$ So $NW_1$ yields the higher payoff.
And now we've run out of things the other player can do, and in every case $NW_1$ yields a higher payoff to player $1$ than $W_1$ does.
So $NW_1$ is dominant.
Player $2$ faces a similar situation; in each case $NW_2$ pays better than $W_2.$
Note that it is not sufficient to observe that the sum of payoffs of $NW_j$ is greater than the sum of payoffs of $W_j$ for player $j$.
Suppose we had another game with this payoff matrix:
$$
\begin{array}{cc}
& \text{Player $2$} \\
\text{Player $1$} &
\begin{array}{c|c|c|}
& A & B \\ \hline
 A & (6,6) & (5,5) \\ \hline
 B & (5,5) & (17,17) \\ \hline
\end{array}
\end{array}
$$
Here, for each player the sum of payoffs for action $B$ is $22$
while the sum of payoffs for action $A$ is only $11,$
but $B$ does not dominate $A$; the payoff for $B$ is less than $A$
when the other player chooses $A.$
