I have recently started to learn about simplices, simplicial complexes and simplicial homology. I understand that for some set of points $\{ p_0, \ldots, p_k \} \subset \mathbb{R}^n$ a simplex $\sigma$ is the set
$$\sigma \enspace \equiv \enspace < \; p_0, \ldots, p_k \; > \enspace := \enspace \Big\{ \; x \in \mathbb{R}^n \; \Big | \; x = \sum_{i = 0}^k \lambda_i p_i \, , \; \lambda_i > 0 \, , \; \sum_{i = 0}^k \lambda_i = 1 \; \Big\}$$
Now, the author of my script started adding and substracting simplices, e.g. when defining the boundary of $\sigma$ he writes
$$\partial_k \sigma \enspace = \enspace < \; p_1, \ldots, p_k \; > - < \; p_0, p_2, \ldots, p_k \; > + \ldots + < \; p_0, \ldots, p_{k-1} \; >$$
However, he did not define what the operations $+$ and $-$ mean in this context. I also seem to have trouble finding a rigorous definition of those operations for simplices on the web. Can anyone explain?
The short answer is: the maps $\partial$ are defined on the free abelian groups generated by $n$-simplices, hence it makes sense to speak about sums of simplices in these groups.
I'll try to give a down to earth answer, for which we will need to detour first. Consider the set $\mathbb{Z}[X]$ of integer polynomials. An element there is an expression $$ a_0 + a_1 X + \ldots + a_nX^n $$ where each $a_i$ is an integer, and $X$ is an indeterminate. Even though we might be quite familiar with thinking of polynomials in this way, the terms 'expression' and 'indeterminate' are informal. A rigorous definition of $\mathbb{Z}[X]$ could be the set of sequences of integers that are eventually zero,
$$ \mathbb{Z}[X] = \{(a_n)_{n \geq 0} \in \mathbb{Z}^{\mathbb{N}_0} : \text{there exists $k \in \mathbb{N}_0$ such that $a_n = 0$ for all $k \geq n$}\}. \tag{1} $$
Given integers $a_1,\ldots,a_n\in \mathbb{Z}$, we then define
$$ a_0 + a_1 X + \ldots + a_nX^n := (a_0,a_1,\ldots,a_n,0,0,\ldots) \tag{2}. $$
All the operations on polynomials can be defined in terms of sequences as in $(1)$ and are compatible with the usual notation $(2)$. This is a way to formally capture the idea of having some unknown element $X$ that can be 'scaled' by integers, multiplied by itself, and such that we can make sense of the sums of expressions like this.
In the same spirit, suppose you have a set $X$ and you want to have another set $M$ such that:
A way to do this is to define the free abelian group $\mathbb{Z}^{(X)}$ with basis $X$. This concept is closely related to bases of vector spaces. The definition (technically, one definition, there are 'many' equivalent ones) is as follows: as a set $\mathbb{Z}^{(X)}$ consists of functions $f \colon X \to \mathbb{Z}$ such that $f(x) \neq 0$ for finitely many $x \in X$. In other words, the elements of $\mathbb{Z}^{(X)}$ are finitely supported functions $\mathbb{Z} \to X$. We can make sense of sum and multiplication of elements here in the same way we do to define a vector space structure on $\mathbb{R}^X$. Namely:
You can check that these operations define once again elements of $\mathbb{Z}^{(X)}$.
Now, let's go back to the analogy with polynomials; in particular, to the relation between definition $(1)$ and notation $(2)$. Even though formally the free abelian group is defined as finitely supported functions, we want to think of it as "$\mathbb{Z}$-linear combinations of $X$". To do this, we write $$ x := \chi_{\{x\}}, \quad \chi_{\{x\}}(y) = \begin{cases}1 &\text{if $x=y$}\\ 0 &\text{otherwise}\end{cases} $$ You can check that, following this notation, every element of $\mathbb{Z}^{(X)}$ can be written as a finite sum $$ a_1 x_1 + \ldots +a_n x_n $$ for some $x_i \in X$ and integers $a_i$. Moreover, two such expressions are equal if the elements of $X$ appearing are the same, and the coefficients accompanying each element coincide.
In this case, we consider $\mathbb{Z}^{(C_n)}$ with $C_n$ the set of $n$-simplices. In $\mathbb{Z}^{(C_n)}$ it makes perfect sense to speak of $3 \sigma + 22 \tau$ for some pair of simplices $\sigma,\tau$ or any expression of this sort; likewise the maps $\partial_k$ are well defined.
I hope this gives a little bit more context to the first sentence of this answer.