The Cauchy functional equation asks about functions $f \colon \mathbb R \to \mathbb R$ such that
$$f(x+y)=f(x)+f(y).$$
It is a very wellknown functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called additive functions.
Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.)
Is there some overview of basic facts about Cauchy equation and related functional equations – preferably available online?
I have no doubt that many resources about Cauchy functional equations and its relatives are available. But many properties of them have been shown in MSE posts, I will try to provide here list of links to the questions I am aware of. I have made this post CW, feel free to add more links and improve this answer in any way. Several links have been collected also in this post: Functional Equation $f(x+y)=f(x)+f(y)+f(x)f(y)$
$\newcommand{\R}{\mathbb{R}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}$
Cauchy functional equation
We are interested in functions $\Zobr f{\R}{\R}$ such that
$$f(x+y)=f(x)+f(y) \tag{C}$$
holds for each $x,y\in\R$.
 If $f$ is a solution of (C) and $q$ is a rational number, then $f(qx)=qf(x)$ holds for each $x\in\R$.
See e.g. If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$
 Every continuous solution $\Zobr f{\R}{\R}$ of (C) has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. this question: I want to show that $f(x)=x.f(1)$ where $f:R\to R$ is additive. or I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$
 If $\Zobr f{\R}{\R}$ is a solution of (C) which is continuous at some point $x_0$, then it is continuous everywhere.
See e.g. Proving that an additive function $f$ is continuous if it is continuous at a single point
 Every Lebesgue measurable solution $\Zobr f{\R}{\R}$ of (C) has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable or Measurable Cauchy Function is Continuous or Prove that if a particular function is measurable, then its image is a rect line or Additivity + Measurability $\implies$ Continuity.
 If a solution $f$ of (C) is bounded on some interval, then $f(x)=cx$ for some $c\in\R$.
One part of this question is about the proof that locally bounded solution of (C) is necessarily continuous: Real Analysis Proofs: Additive Functions
 Every integrable solution $\Zobr f{\R}{\R}$ of (C) has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable
 There exist noncontinuous solutions of (C).
See noncontinuous function satisfies $f(x+y)=f(x)+f(y)$,
Do there exist functions satisfying $f(x+y)=f(x)+f(y)$ that aren’t linear?,
On sortoflinear functions, Many other solutions of the Cauchy’s Functional Equation
NOTE: The proof of this fact needs at least some form of Axiom of Choice – it is not provable in ZF.
For the role of AC in this result, see this MO thread, and theses questions: Cauchy functional equation with non choice, A question concerning on the axiom of choice and Cauchy functional equation, Is there a nontrivial example of a $\mathbb Q$endomorphism of $\mathbb R$?, What is $\operatorname{Aut}(\mathbb{R},+)$?
 The graph $G(f)=\{(x,f(x)); x\in\R\}$ is a dense subset of $\R^2$ for every noncontinuous solution of (C).
See e.g. Graph of discontinuous linear function is dense
 If we have the equation (C) for function $\Zobr f{[0,\infty)}{[0,\infty)}$, then every solution is of the form $f(x)=cx$.
See If $f:[0,\infty)\to [0,\infty)$ and $f(x+y)=f(x)+f(y)$ then prove that $f(x)=ax$
Related equations
There are several functional equations that are closely related to Cauchy equation. They can be reduced to it by some methods or solved by very similar methods as the Cauchy equation.
The equation $f(x+y)=f(x)f(y)$
$$f(x+y)=f(x)f(y) \tag{1}$$
 If $f$ is a solution of (1), then either $f=0$ (i.e., $f$ is constant function equal to zero) or $f(x)>0$ for each $x\in\R$.
See e.g. Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$? and A nonzero function satisfying $g(x+y) = g(x)g(y)$ must be positive everywhere

If $f$ is a solution of (1), $x\in\Q$ and if we denote $a=f(1)$, then $f(x)=a^x$.

Every continuous solution of (1) has the form $f(x)=a^x$ for some $a\ge 0$.
See e.g. continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$ or How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$, assuming f is continuous and not zero?
 If a solution of (1) is continuous at $0$, then it is continuous everywhere.
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere
 There are noncontinuous solutions of (1).
They can be obtained from noncontinuous solutions of (C).
See Is the group isomorphism $\exp(\alpha x)$ from the group $(\mathbb{R},+)$ to $(\mathbb{R}_{>0},\times)$ unique?
 If a solution of (1) is differentiable at $0$, then it is differentiable everywhere and $f'(x)=f(x)f'(0)$.
See Prove that $f’$ exists for all $x$ in $R$ if $f(x+y)=f(x)f(y)$ and $f'(0)$ exists and Differentiability of $f(x+y) = f(x)f(y)$ There are also posts searching for differentiable solutions: Solution for exponential function’s functional equation by using a definition of derivative
The equation $f(xy)=f(x)+f(y)$
$$f(xy)=f(x)+f(y) \tag{2}$$
Examples of functions where $f(ab)=f(a)+f(b)$
The equation $f(xy)=f(x)f(y)$
$$f(xy)=f(x)f(y) \tag{3}$$
 Every continuous solution of (3) has the form $f(x)=x^a$
See If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
 A solution of (3) which is continuous at $1$ is continuous for each $x>0$.
See Functional equation $f(xy)=f(x)+f(y)$ and continuity
 If a function fulfills both (C) and (3), then either $f(x)=0$ or $f(x)=x$ (i.e., it is either zero function or the identity).
Notice that we do not require continuity here. See Functional equations $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$