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.
The statement "since it is well-defined, it is unique" is certainly misleading.
Uniqueness is shown (independently from existence = being well-defined) as follows: Assume $U\colon V/W\to Z$ is another linear map such that also $U(\pi(x))=T(x)$. Then $U=\tilde T$ because for any $y\in V/W$ there exists $x\in V$ such that $y=\pi(x)$ and then $U(y)=U(\pi(x))=T(x)=\tilde T(\pi(x))=\tilde T(y)$.
Note specifically that we did not use any properties of $T$ for uniqueness, whereas well-definedness does require some additional properties ...
Best Answer
The thing we need to prove is as follows: if $S$ is a fixed subspace, and if $P,P'$ are two projections onto $S$, then $P = P'$.
An easy way to prove this is as follows: we note that $P,P'$ must be identical if there is some basis $e_1,\dots,e_n$ for $\Bbb R^n$ such that $P e_i = P' e_i$ for all $1 \leq i \leq n$. We construct such a basis as follows:
We may select an orthonormal basis $e_1,\dots,e_r$ of $S$. We may extend this to an orthogonal basis of $\Bbb R^n$. We then note that we must have $$ P(e_i) = P'(e_i) = \begin{cases} e_i & 1 \leq i \leq r\\ 0 & r < i \leq n\end{cases} $$ The conclusion follows.