Let me refresh you with some results.
Recall that, as a simple consequence of the definition of inner product, the brackets $\langle \cdot,\cdot \rangle$ take out the scalars from the first entry, but they bring out the complex conjugates from the second, that is,
$$\langle u,cv \rangle = \overline{c} \langle u,v \rangle. \tag{1}$$
Also, following the analogy that in $\mathbb{R}^3$ (to be illustrative, it doesn't really matter) every vector $\mathbf{x} = (x,y,z)$ can be written as
\begin{align}
\mathbf{x} &= xe_1 + ye_2 + ze_3 \\
&= \langle \mathbf{x},e_1 \rangle e_1 + \langle \mathbf{x},e_2 \rangle e_2 + \langle \mathbf{x},e_3 \rangle e_3
\end{align}
where $e_1$, $e_2$ and $e_3$ are the standard basis vectors, then, any vector $v$ in an abstract inner product space can be written as
$$v = \langle v,w_1 \rangle w_1 + \langle v,w_2 \rangle w_2 + \cdots + \langle v,w_n \rangle w_n \tag{2}$$
where $w_1,w_2,\dots,w_n$ must be form an orthonormal basis for the whole space (which is the case for our example in $\mathbb{R}^3$).
Now, we start with the given problem. We want to find some vector $y \in \textsf{V}$ such that $g(x) = \langle x,y \rangle$ for all $x \in \textsf{V}$, that is, the rule of correspondence of $g$ is completely determined by $y$.
First, choose an orthonormal basis for $\textsf{V}$, let's say, $v_1,v_2,\dots,v_n$ (this can be done since the space is finite-dimensional). Then, for any $x \in \textsf{V}$ we have
\begin{align}
g(x) &= g(\langle x,v_1 \rangle v_1 + \langle x,v_2 \rangle v_2 + \cdots + \langle x,v_n \rangle v_n) \tag{3} \\
&= \langle x,v_1 \rangle g(v_1) + \langle x,v_2 \rangle g(v_2) + \cdots + \langle x,v_n \rangle g(v_n) \tag{4} \\
&= \langle x,\overline{g(v_1)}v_1 \rangle + \langle x,\overline{g(v_2)}v_2 \rangle + \cdots + \langle x,\overline{g(v_n)}v_n \rangle \tag{5} \\
&= \langle x, \overline{g(v_1)}v_1 + \overline{g(v_2)}v_2 + \cdots + \overline{g(v_n)}v_n \rangle \tag{6}
\end{align}
where $(3)$ happens by our observation $(2)$, $(4)$ by the linearity of $g$, $(5)$ by $(1)$ since $g(v_1),\dots,g(v_n)$ are scalars, and $(6)$ follows from the linearity in the second entry of the inner product.
So, letting $y := \overline{g(v_1)}v_1 + \overline{g(v_2)}v_2 + \cdots + \overline{g(v_n)}v_n$ in the last step we have that $g(x) = \langle x,y \rangle$, as desired. To show the uniqueness, suppose there are $y'$ such that $g(x) = \langle x,y' \rangle$ for any $x$. Then, for any $x \in \textsf{V}$,
$$\langle x,y-y' \rangle = \langle x,y \rangle - \langle x,y' \rangle = 0.$$
In particular, for $x=y-y'$ we see that $\langle y-y',y-y' \rangle = 0$, which only happens when $y-y'=0$, that is, $y'$ must be the same as $y$.