In extended form the inequality constraints in first case can be written as
$$a_{1,1}x_1+a_{1,2}x_2+...+a_{1,n}x_n\le b_1$$
$$a_{2,1}x_1+a_{2,2}x_2+...+a_{2,n}x_n\le b_2$$
$$...$$
$$a_{m,1}x_1+a_{m,2}x_2+...+a_{m,n}x_n\le b_m$$
Let's introduce some slack variables $x_{n+i}\ge 0$ into the inequality constraints such that
$$a_{1,1}x_1+a_{1,2}x_2+...+a_{1,n}x_n+x_{n+1}= b_1$$
$$a_{2,1}x_1+a_{2,2}x_2+...+a_{2,n}x_n+x_{n+2}= b_2$$
$$...$$
$$a_{m,1}x_1+a_{m,2}x_2+...+a_{m,n}x_n+x_{n+m}= b_m$$
In short form
$$\sum_{j=1}^{n} a_{i,j}x_j+x_{n+i}=b_i\quad for\,i=1,...,m$$
And the first case is transformed into the second one
------EDIT-------
We can rewrite the equality conditions in B such as
$$\sum_{j=1}^{n} a_{i,j}x_j=b_i-x_{n+i}\quad for\,i=1,...,m$$
Since the question states that $x_j\ge 0$ for $j=1,...,m+n$ we can eliminate $x_j$ for $j> n$ by using inequality constraints without loss of generality such that
$$\sum_{j=1}^{n} a_{i,j}x_j\le b_i\quad for\,i=1,...,m$$
- $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.
- $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
\begin{align}
4y_1 -y_2 = 7 > -7 &\implies [y_1, y_2]^T \text{ infeasible} \\
4y_1 -y_2 = 7 \ne -7 &\implies [y_1, y_2]^T \text{ non-basic}
\end{align}
The second implication follows since we actually transform the LP into
\begin{array}{lrr}
\text{max} & w = -22y_1 -8y_2 &\\
\text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \\
& 4y_1 -y_2 +s_2 =& -7\\
&-y_1 -3y_2 +s_3 =& 5\\
&8y_1 -y_2 +s_4 =& -14\\
& y_1, y_2, s_1, s_2, s_3, s_4 \geq 0
\end{array}
The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.
- You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,\dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.
Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 \ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.
Current basis: $s_1, s_2, s_3, s_4$
\begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \\ \hline
s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \\
s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \\
s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \\
s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \\ \hline
& 22 & 8 & 0 & 0 & 0 & 0 & 0 \\
\text{ratio} & 11/4 & -8 & & & & 0 &
\end{array}
Leaving variable: $s_4$, entering variable: $y_2$
Current basis: $s_1, s_2, s_3, y_2$
\begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \\ \hline
s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \\
s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \\
s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \\
y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \\ \hline
& \color{blue}{86} & \color{blue}{0} & \color{red}{0} & \color{red}{0} & \color{red}{0} & \color{red}{\bbox[2px, border: solid 1px]{8}} & -112 \\
\end{array}
Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.
By adding slack variabes $t_1,t_2$ to the primal
\begin{array}{rlr}
\min z = & \color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\\
\text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +\color{blue}{t_1} &= 22 \\
& x_1 + x_2 + 3x_3 + x_4 +\color{blue}{t_2} &= 8 \\
& \color{red}{x_1, x_2, x_3, \bbox[2px, border: solid 1px]{x_4}}, \color{blue}{t_1,t_2} \geq 0,
\end{array}
we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[\color{red}{x_1, x_2, x_3, \bbox[2px, border: solid 1px]{x_4}}, \color{blue}{t_1,t_2}]^T = [\color{red}{0,0,0, \bbox[2px, border: solid 1px]{8}}, \color{blue}{86,0}]$.
Best Answer
I think there are some details missing in your question.
Let the primal be the minimization of $c^Tx$, subject to $Ax = b, x\geq 0$. We will show the following: if the optimal simplex tableau gives us a non-degenerate basic feasible solution with $c_i - z_i \geq 0$ for all variables, then the dual has a unique optimal solution.
Use the complementary slackness conditions.
Note that rank$(A^T_{n \times m})$ = rank$(A_{m \times n}) = m$, a common assumption (because we can just delete rows if redundant). This means the basis matrix $B$ is of dimension $m \times m$.
There are $m$ constraints, and so, $m$ dual variables $y_1, \ldots y_m$.
Consider the basic variable $x_{Bi}$ in this optimal BFS $x$, and an optimal solution $y$ of the dual. As the BFS is non-degenerate, $x_{Bi} > 0$. The complementary slackness conditions give, $x_{Bi} (A^Ty - c)_{Bi} = 0$ which imply, $(A^Ty)_{Bi} = a_{Bi}^Ty = c_{Bi}$. Here, $a_{Bi}^T$ is the row corresponding to the basic variable $x_{Bi}$ in $A^T$. Repeating this for all the $m$ basic variables, we get the system of equations, compactly represented as $(A^Ty)_{B} = c_{B}$. By the definition of a basis, the set of all rows $a_{B1}^T, \ldots, a_{Bm}^T$ are all linearly independent. Thus, there exists only a unique solution $y$ to $(A^Ty)_{B} = c_{B}$.
In fact, multiplying by $(B^{-1})^T$ gives us $y = (B^{-1})^Tc_{B}$ as claimed. Thus, this solution is unique.
We must also ensure that $y$ is feasible for the dual. This comes from the fact that $c_i - z_i = c_i - a_i^Ty \geq 0$ for all $i$. The dual's feasible region is given by $A^Ty \leq c$. So, $y$ is feasible as well. This is what we had to show!