Claim: There exists a function $f : \mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$, but which is not linear.
Proof of claim: To prove the existence of $f$, define an equivalence relation $\sim$ on $\mathbb{R}$ by
$$
x \sim y \mbox{ iff } y-x \in\mathbb{Q}.
$$
This defines equivalence classes $[x] = \{ y \in \mathbb{R} : y \sim x \}$.
As with all equivalence relations, either $[x]\cap[y] = \emptyset$ or $[x]=[y]$. The collection of equivalence classes becomes a vector space over the field of rational numbers $\mathbb{Q}$ under the operations
$$
[x]+[y]=[x+y],\;\;\; \alpha[x]=[\alpha x],\;\;\;\alpha\in\mathbb{Q},\;x,y\in\mathbb{R}.
$$
These linear operations are well-defined because $x\sim x'$, $y\sim y'$ and $\alpha\in\mathbb{Q}$ give
$$
x+y \sim x'+y',\;\;\; \alpha x \sim \alpha x'.
$$
So the space $X$ consisting of all equivalence classes $[x]$ is a linear space over the field of rational numbers. Hence, $X$ has a Hamel basis $\{ [x_{\alpha}]\}_{\alpha\in\Lambda}$. Now, for any choice function $\{s_{\alpha}\}_{\alpha\in S}\subseteq \mathbb{R}$ from $\Lambda$ to $\mathbb{R}$, there is a unique function $F : X\rightarrow\mathbb{R}$ that is linear over $\mathbb{Q}$ and satisfies:
$$
F([x_{\alpha}]) = s_{\alpha}.
$$
The function $f(x) = F([x])$ is additive and is linear over $\mathbb{Q}$, but you can choose $\{ s_{\alpha} \}$ so that $f$ is not linear. $\;\;\Box$
Quadratic Form from an Additive Function: For any additive function $f : \mathbb{R}\rightarrow\mathbb{R}$, the following defines a quadratic form
$$
q(x) = (f(x))^{2}.
$$
This is easily checked:
\begin{align}
q(x+y)+q(x-y) & = (f(x)+f(y))^{2}+(f(x)-f(y))^{2}\\
& = 2f(x)^{2}+2f(y)^{2} = 2q(x)+2q(y).
\end{align}
An additive function on $\mathbb{R}$ is always linear over the rationals $\mathbb{Q}$. But without some kind of measurability or continuity, the function $f$ may not be linear over $\mathbb{R}$, which is equivalent to continuity over $\mathbb{R}$.
Theorem: An additive function $f : \mathbb{R}\rightarrow \mathbb{R}$ is linear over $\mathbb{R}$ iff it is Lebesgue measurable. Equivalently, $f$ is continuous.
What makes a norm different?: There is a classic theorem which states that a norm is generated by an inner product iff the norm satisfies the parallelogram law. The proof generally starts by showing that, if $\|\cdot\|$ satisfies the Parallelogram Law, the function
$$
\rho(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2}
$$
is additive in both coordinates $x$, $y$. Then it follows that $\rho(q x,y)=\rho(x,qy) = q\rho(x,y)$ for all $q \in \mathbb{Q}$. If you're dealing with a complex space, then $\rho(ix,y)=i\rho(x,y)$ and $\rho(x,iy)=-i\rho(x,y)$ follow from $\|ix\|=\|x\|$. What makes the norm different is that $\rho(x,x) \ge 0$; hence, the Cauchy-Schwarz inequality holds:
$$
|\rho(x,y)|^{2} \le \rho(x,x)\rho(y,y) = \|x\|^{2}\|y\|^{2}
$$
From this you get continuity of $\rho(x,y)$ with respect to the norm. So, for real $r$ and rational $\{ q_{n} \}$ converging to $r$,
$$
\begin{align}
|\rho(rx,y)-q_{n}\rho(x,y)|
& \le \|rx-q_{n}x\|\|y\| \\
& =|r-q_{n}|\|x\|\|y\|\rightarrow 0.
\end{align}
$$
And that's enough to guarantee that $\rho(rx,y)=r\rho(x,ry)$ for all $r\in\mathbb{R}$.
The result below is useful for proving that a quadratic form arises from a sesquilinear form.
Theorem: [Bounded Quadratic Form Representation] Let $Q$ be a quadratic form on a complex linear space $X$. If there exists a seminorm $|\cdot|$ on $X$ such that $|Q(x)| \le |x|^{2}$, then $Q(x)= S(x,x)$ for a unique sesquilinear form $S(\cdot,\cdot)$ on $X\times X$.
