If $f \in H^*$, there is a unique $y \in H$ such that $f(x) = \langle x, y \rangle$ for all $x \in H$. This is from a theorem in Folland, where he is showing that the map $y \mapsto f_y$ where $f_y(x) = \langle x, y \rangle$ is a conjugate-linear isometry of $H$ into $H^*$, where $H$ is a Hilbert space.
The theorem itself is:

If $f \in H^*$, there is a unique $y \in H$ such that $f(x) = \langle x, y \rangle$ for all $x \in H$.

His proof is as follows:

If $\langle x, y \rangle = \langle x, y'\rangle$ for all $x$, by taking $x = y - y'$ we conclude that $\|y - y'\|^2 = 0$ and hence $y = y'$. If $f$ is the zero functional, then obviously $y = 0$. Otherwise, let $M = \{x \in H : f(x) = 0\}$. Then $M$ is a proper closed subspace of $X$, so $M^\perp \neq \{0\}$ by Theorem 5.24. Pick $z \in M^\perp$ with $\|z\| = 1$. If $u = f(x)z - f(z)x$ then $u \in M$, so
$$ 0 = \langle u, z \rangle = f(x)\|z\|^2 - f(z) \langle x,z \rangle = f(x) - \langle x, \overline{f(z)}z \rangle.$$
Hence, $f(x) = \langle x, y \rangle$ where $y = \overline{f(z)}z$.

I have a few questions regarding his proof. First, how does it follow that $u \in M$? He never specified what the element $x$ is here, but regardless, by the linearity of $f$ the image of $u$ is:
$$f(u) = f(x)f(z) - f(z)f(x).$$
How is the above equal to 0, and why does Folland choose $\|z\| = 1$?
Furthermore, how does it follow that $\langle u, z \rangle = f(x)\|z\|^2 - f(z) \langle x,z \rangle = f(x) - \langle x, \overline{f(z)}z \rangle$?
Taking a step back and looking at the bigger picture, how exactly does this proof show surjectivity?
Maybe I have misunderstood the theorem and the relationship between $H$ and its dual. My current understanding is that, since $H$ and $H^*$ is isomorphic, clearly each $y$ corresponds to some $f_y$, and the theorem tells us just how exactly $f_y$ should be defined. Is this correct?
 A: In the definition of $u$, $x$ is an arbitrary element of $H$; we are going to show that $f(x)=\langle x,y\rangle$ for a $y$ to be specified shortly. The vector $u$ depends on $x$. It is only an auxiliary notation for the computation.
Why is $u$ in $M$? As you note, $f(u) = f(x)f(z) - f(z)f(x)$. Why is this $0$? Because $f$ takes values in the field over which $H$ is defined, and so multiplication is commutative! $f(x)f(z)=f(z)f(x)$.
Folland picks $z$ with $\lVert z\rVert = 1$ for the next calculation. We have:
$$\begin{align*}
0 &= \langle u,z\rangle &\text{(because }u\in M\text{ and }z\in M^{\perp}\text{)}\\
&=\langle f(x)z-f(z)x,z\rangle &\text{(by definition of }u\text{)}\\
&= f(x)\langle z,z\rangle - f(z)\langle x,z\rangle &\text{(by linearity of the inner product)}\\
&= f(x)\lVert z\rVert^2 - f(z)\langle x,z\rangle\\
&= f(x) - f(z)\langle x,z\rangle &\text{(because we picked }z\text{ with }\lVert z\rVert=1\text{)}\\
&= f(x) - \langle x,\overline{f(z)}z\rangle &\text{(by conjugate homogeneity)}
\end{align*}$$
This shows that $f(x) = \langle x,\overline{f(z)}z\rangle$ for each $x$. Note that even though $u$ depended on $x$, $z$ does not. The vector $u$ was an auxiliary to the computation.
This shows that $f=f_{y}$ with $y=\overline{f(z)}z$. How does this prove surjectivity? We have a map from $H$ to $H^*$ given by $y\longmapsto f_y$. We have now shown that every element of $H^*$ is of the form $f_y$ for some $y$, which shows this map is surjective. Note that what we proved was that given any $f\in H^*$, there exists a $y\in H$ such that $f=f_y$.  The uniqueness clause, on the other hand, shows that the map $y\longmapsto f_y$ is one-to-one. Thus, you have a bijection.
