Let's take a step back, and look at it from a more abstract perspective.
First, we define a map $\lambda \colon W \to V^\ast$ by
$$\lambda(w) \colon v \mapsto \langle f(v), w\rangle_W,$$
or $\lambda(w) = \langle\,\cdot\,,w\rangle_W \circ f$. Now, looking at it, we notice that $\lambda$ is linear (or, if the scalar field is $\mathbb{C}$, and the inner product linear in the first and antilinear in the second argument, $\lambda$ is antilinear; if the spaces are complex and the inner product is antilinear in the first, and linear in the second argument, consider $\langle w,f(v)\rangle_W$ instead):
$$\begin{align}
\lambda(w_1+w_2)(v) &= \langle f(v), w_1 + w_2\rangle_W\\
&= \langle f(v), w_1\rangle_W + \langle f(v),w_2\rangle_W\\
&= \lambda(w_1)(v) + \lambda(w_2)(v)\\
&= \bigl(\lambda(w_1) + \lambda(w_2)\bigr)(v),
\end{align}$$
and since $v$ was arbitrary, that means $\lambda(w_1+w_2) = \lambda(w_1) + \lambda(w_2)$. And for a scalar $s$, we have
$$\begin{align}
\lambda(s\cdot w)(v) &= \langle f(v), s\cdot w\rangle_W\\
&= s\cdot \langle f(v), w\rangle_W\\
&= s\cdot \lambda(w)(v)\\
&= \bigl(s\cdot \lambda(w)\bigr)(v),
\end{align}$$
that is, $\lambda(s\cdot w) = s\cdot \lambda(w)$ (for complex scalars, we would have to conjugate $s$ and $\lambda$ would be antilinear).
Now, with the (anti-) isomorphism $\sigma \colon V \to V^\ast$, given by $\sigma(v) = \langle\,\cdot\,,v\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle w, v\rangle_V$ (or $\sigma(v) = \langle v,\,\cdot\,\rangle_V$, i.e. $\sigma(v) \colon w \mapsto \langle v,w\rangle_V$ if $V$ is a complex vector space and the inner product is antilinear in the first argument, as is customary in physics), consider the composition
$$\varphi = \sigma^{-1} \circ \lambda \colon W \to V.$$
Then $\varphi$ is linear (as the composition of two linear maps in the real case, and as the composition of two antilinear maps in the complex case), and for $v\in V,\; w \in W$, we have
$$\begin{align}
\langle v, \varphi(w)\rangle_V &= \langle v, \sigma^{-1}(\lambda(w))\rangle_V\\
&= \lambda(w)(v)\tag{definition of $\sigma$}\\
&= \langle f(v),w\rangle_W,\tag{definition of $\lambda$}
\end{align}$$
and that means we can define $f^\ast := \varphi = \sigma^{-1}\circ \lambda$.
It remains to be seen that there is only one such map $W \to V$.
Suppose that $\psi \colon W \to V$ is a map with $\langle f(v), w\rangle_W = \langle v, \psi(w)\rangle_V$ for all $v\in V,\; w\in W$.
Fix an arbitrary $w_0\in W$. Then for all $v\in V$
$$\begin{align}
\langle v,\psi(w_0) - f^\ast(w_0)\rangle_V &= \langle v,\psi(w_0)\rangle_V - \langle v, f^\ast(w_0)\rangle_V\\
&= \langle f(v), w_0\rangle_W - \langle f(v), w_0\rangle_W\\
&= 0.
\end{align}$$
In particular, we can choose $v = \psi(w_0) - f^\ast(w_0)$, so we have
$$\langle \psi(w_0) - f^\ast(w_0), \psi(w_0) - f^\ast(w_0)\rangle_V = 0,$$
which means $\psi(w_0) - f^\ast(w_0) = 0$. Since $w_0 \in W$ was arbitrary, it follows that $\psi = f^\ast$, and the uniqueness is also established.
If $\langle x, y \rangle = 0$ for all $x$, then we must have $y=0$. To see this, just choose $x=y$ which gives $\langle y, y \rangle = \|y\|^2 =0$ from which it follows that $y = 0$.
So, if $\langle v, w' \rangle = \langle v, w'' \rangle$ for all $v$, then you have $\langle v, w'-w'' \rangle = 0 $ for all $v$ from which it follows that $w'=w''$.
Existence is straightforward to establish if you have an orthonormal basis, say $e_k$:
Then, with $v= \sum_k v_k e_k$, we have
$\langle f(v), w \rangle = \sum_k \overline{v_k} \langle f(e_k), w \rangle =
\langle \sum_k v_k e_k, \sum_k \langle f(e_k), w \rangle e_k \rangle$, that is, $\langle f(v), w \rangle = \langle v, w' \rangle $, where
$w'=f^*(w) = \sum_k \langle f(e_k), w \rangle e_k $.
It is straightforward to see that the function $ w \mapsto
\sum_k \langle f(e_k), w \rangle e_k $ is linear
and the comment above shows that the value $f^*(w)$ is unique, from which it follows that $w \mapsto f^*(w)$ is unique.
Best Answer
You are given $E$ and $F$ to be Euclidean spaces, with inner products $\langle\cdot,\cdot\rangle_E :E\rightarrow \mathbb{R}$ and $\langle\cdot,\cdot\rangle_F :F\rightarrow \mathbb{R}$. Note that the linear map $T:u \mapsto \langle f(u), v \rangle_F$ is uniquely defined by $v\in F$ and $f$. Since $T$ is a linear functional on $E$, hence there exist a unique vector $w$ in $E$, such that $T(u)=\langle u, w\rangle_E$, i.e. $$\langle f(u), v \rangle_F=\langle u, w\rangle_E, $$let's call $w=f'(v)$, since this vector is associated to $v$, while defining $T$. It is required be to check that $f':F\rightarrow E$ is a linear map. And, then we have $$\langle u, f'(v)\rangle_E = \langle f(u), v \rangle_F.$$ I guess the notation, you have provided should be $\Psi_E:E\rightarrow E^*$ with $\Psi_E(u)=\langle \cdot, u\rangle_E$, we get $$\Psi_E(f'(v))(u)= \langle u, f'(v)\rangle_E = \langle f(u), v \rangle_F.$$