Let $T \colon \mathbb{C}^n \to \mathbb{C}^n$ be a linear operator, which is bounded of course; so the Hilbert adjoint operator $T^* \colon \mathbb{C}^n \to \mathbb{C}^n$ is a well-defined bounded linear operator with $\Vert T^* \Vert = \Vert T \Vert$.
Suppose that $A$ and $B$ are the matrices of $T$ and $T^*$, respectively, with respect to a given basis for $\mathbb{C}^n$.
Then how do we obtain the following relations? $$\langle Tx, y \rangle = \left( Ax \right)^T \overline{y} \ \mbox{ for all } \ x, y \in \mathbb{C}^n.$$ And, $$\langle x, T^*y \rangle = x^T\overline{By} \ \mbox{ for all } \ x, y \in \mathbb{C}^n.$$
Here we treat any $u \in \mathbb{C}^n$ as a coluumn vector. Moreover, $u^T$ is the row vector that is the transpose of $u$, and $\overline{u}$ is the column vector obtained from $u$ by replacing each component of $u$ by its complex conjugate.
After @Martin Argerami's comment.
Let $\{e_1, \ldots, e_n\}$ be an ordered basis for $\mathbb{C}^n$.
Then each $x \in \mathbb{C}^n$ has a unique representation $$x = \sum_{j=1}^n \xi_j e_j,$$ where $\xi_1, \ldots, \xi_n \in \mathbb{C}$.
For each $j = 1, \ldots, n$, since $T e_j \in \mathbb{C}^n$, there are scalars $\alpha_{ij} \in \mathbb{C}$ for each $i = 1, \ldots, n$ such that $$T e_j = \sum_{i=1}^n \alpha_{ij} e_i.$$
Since $Tx \in \mathbb{C}^n$, we have $$T x = \sum_{i=1}^n \eta_i e_i \ \tag{equation (1) } $$ for a unique $n$-tuple of complex numbers $\eta_1, \ldots, \eta_n$.
But since $x = \sum_{j=1}^n \xi_j e_j$ and since $T$ is linear, we have $$ \begin{align*} Tx & = T( \sum_{j=1}^n \xi_j e_j ) \\ & = \sum_{j=1}^n \xi_j T e_j \\ &= \sum_{j=1}^n ( \xi_j \sum_{i=1}^n \alpha_{ij} e_i ) \\ & = \sum_{j=1}^n \sum_{i=1}^n \xi_j \alpha_{ij} e_i \\ & = \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i. \ \tag{equation (2)} \end{align*}$$
From equation (1) and (2), we obtain $$ \sum_{i=1}^n \eta_i e_i = Tx = \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i. $$ So $$ \sum_{i=1}^n \eta_i e_i - \sum_{i=1}^n (\sum_{j=1}^n \alpha_{ij} \xi_j ) e_i = 0.$$ Or, $$ \sum_{i=1}^n \left( \eta_i - \sum_{j=1}^n \alpha_{ij} \xi_j \right) e_i = 0.$$ Since $e_1, \ldots, e_n$ are linearly independent, we have $$ \eta_i = \sum_{j=1}^n \alpha_{ij} \xi_j \ \mbox{ for each } \ i = 1, \ldots, n. \tag{equation (3)} $$
The matrix $$ A \colon= [\alpha_{ij} ]_{m \times n}$$ is the matrix of the operator $T$ with respcet to the given basis $\{ e_1, \ldots, e_n \}$.
The column vector $$\tilde{x} \colon= [\xi_1, \ldots, \xi_n ]^T$$ is the co-ordinate vector of $x$ with respect to this basis.
And, the column vector $$\tilde{Tx} \colon= [\eta_1, \ldots, \eta_n ]^T$$ is the co-ordinate vector of $Tx $ with respect to the same basis.
Thus we have $$\tilde{Tx } = A \tilde{x} \ \mbox{ for all } \ x \in \mathbb{C}^n. \ \tag{equation (4)} $$
Now let $y \in \mathbb{C}^n$. Then there are unique complex numbers $\lambda_1, \ldots, \lambda_n$ such that $$y = \sum_{i=1}^n \lambda_i e_i.$$ So we have $$\langle Tx, y \rangle = \langle \sum_{i=1}^n \eta_i e_i, \sum_{i=1}^n \lambda_i e_i \rangle = \sum_{i=1}^n \sum_{j=1}^n \eta_i \overline{\lambda_j} \langle e_i, e_j \rangle.$$
Now what next?
If $\{e_1, \ldots, e_n \}$ is an orthonormal set, then we can say that $$\langle Tx, y \rangle = (Ax)^T \overline{y} = x^T A^T \overline{y}.$$ But what if $\{ e_1, \ldots, e_n \}$ is not orthonormal? How do we show the above equality in that case?
It is simply the equality $$ \langle x,y\rangle =x^T\bar y, $$ together with the fact that $ Tx=Ax $, $T^*x=Bx $.