In the first phase of the simplex method it is convenient to choose the objective function $z=y_1 + \cdots + y_n$ where the $y_i$ variables are the auxiliary variables and $z$ is the objective value of the current tableau(not every textbook writes the $z$ out explicitly). Recall that the auxiliary are introduced to make a starting basis obvious: simply take the set of auxiliary variables as the starting basis. By minimizing this choice of objective if there is a basic feasible solution to the original problem, then this linear program will have an optimal solution with optimal value 0, that is, we can find a solution where all the auxiliary variables are 0. Said another way, we have found values for the $x$ variables such that all the equations are satisfied (without using the $y$'s to pick up any infeasibility).
Now let us look at the mechanics of choosing the initial starting basis to be the set of auxiliary variables $y_1$, $y_2$ and $y_3$. Since these variables are basic, the $x$ variables are non-basic which means $x_1=\cdots=x_5=0$. Thus, $y_1=5$, $y_2=1$ and $y_3=6$.
We can write the objective row as $z=y_1 + y_2 + y_3$, where $z$ is the objective value of the current simplex tableau. The objective row of a simplex tableau is written as $z-y_1-y_2-y_3$ (some books have an explicit column for $z$ which makes this clear). The objective row of a simplex tableau is expressed only in terms of the non-basic variables. Since the objective row is $z-y_1-y_2-y_3$ we simply use the fact that with the current tableau
$$ y_1 = 5 - 2x_1 - x_2 + x_3 $$
$$ y_2 = 1 - - x_2 +x_4$$
$$ y_3 = 6 - 2x_1 - 3x_2 - x_5$$
Substituting this into $z-y_1-y_2-y_3$ we obtain
$$ -12 -4x_1 - 5x_2 + x_3 + x_4 - x_5$$
for the objective row.
The big-M's are used to accelerate driving the auxiliary variables out of the basis. Hopefully this helps.