Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$. On the $\mathbb{K}$-linear space $V$, an inner product is a bilinear form $\varphi \, : \, V \times V \, \longrightarrow \, \mathbb{K}$ which is symmetric positive definite. Let $B = \big( \varepsilon_{1},\ldots, \varepsilon_{n}\big)$ be a basis of $V$ and let $(x,y) \in V^{2}$. We can write :
$$ x = \sum_{i=1}^{n} x_{i} \varepsilon_{i} \quad \mathrm{and} \quad y=\sum_{j=1}^{n} y_{j}\varepsilon_{j} $$
where $(x_{1},\ldots,x_{n}) \in \mathbb{K}^{n}$ and $(y_{1},\ldots,y_{n}) \in \mathbb{K}^{n}$. By bilinearity of $\varphi$, we have :
$$
\begin{align*}
\varphi(x,y) &= {} \varphi \Big( \sum_{i=1}^{n} x_{i} \varepsilon_{i} \, ,\sum_{j=1}^{n} y_{j}\varepsilon_{j} \Big) \\[1mm]
&= \sum_{i,j=1}^{n} x_{i}y_{j} \varphi \big( \varepsilon_{i},\varepsilon_{j} \big)
\end{align*}
$$
As a consequence, in order to define $\varphi$, one must specify the value of $\varphi(\varepsilon_{i},\varepsilon_{j})$ for all $i$ and $j$. Choosing : $\forall i,j, \, \varphi(\varepsilon_{i},\varepsilon_{j}) = \delta_{i,j}$, you obtain an inner product on $V$ for which $B$ is an orthonormal basis.
Let $A$ be a linear transformation where $A(v_i) = e_i$ for each $i$, where $e_1,\dots,e_n$ is the standard basis (verify that such an $A$ exists and must be invertible).
Define the inner product $\langle \cdot,\cdot \rangle_A$ by
$$
\langle x,y \rangle_A = \langle A x,Ay\rangle
$$
verify that $\langle \cdot,\cdot \rangle_A$ satisfies the definition of an inner product.
Best Answer
The real case: for any ${\bf a},{\bf b}$ in $V$ define $$\langle {\bf a},{\bf b}\rangle=\alpha_1\beta_1+\cdots+\alpha_n\beta_n\ ,$$ where ${\bf a}=\alpha_1{\bf v}_1+\cdots+\alpha_n{\bf v}_n$ and ${\bf b}=\beta_1{\bf v}_1+\cdots+\beta_n{\bf v}_n$. In the complex case, $$\langle {\bf a},{\bf b}\rangle=\alpha_1\overline{\beta_1}+\cdots+\alpha_n\overline{\beta_n}$$ with similar notation.
Note that both formulae are well defined, since you have a fixed basis $\{{\bf v}_1,\ldots,{\bf v}_n\}$, so for any ${\bf a}$ there are unique scalars $\alpha_k$ such that ${\bf a}=\alpha_1{\bf v}_1+\cdots+\alpha_n{\bf v}_n$, and similarly for ${\bf b}$.
It is not difficult to prove that these actually are inner products.
Moreover, we have $$\eqalign{ \langle {\bf v}_1,{\bf v}_1\rangle&=1\times1+0\times0+\cdots+0\times0=1\cr \langle {\bf v}_1,{\bf v}_2\rangle&=1\times0+0\times1+\cdots+0\times0=0\cr}$$ and so on, which shows that the basis is orthonormal with respect to this inner product.