For a given space $X$, define $S_1(X)$ to be the free abelian group with basis all paths $\sigma: I \rightarrow X$, and let $S_0(X)$ be the free abelian group with basis $X$.
$(i)$ Show that there is a homomorphism $f: S_1(X) \rightarrow S_0(X)$ with $f(\sigma) = \sigma(1) - \sigma(0)$ for every path in $X$.
What does it mean for $S_1(X)$ to be a free abelian group with basis all paths $\sigma: I \rightarrow X$? Since it's free abelian, $S_1(X) = \sum_{\sigma \text{ $\in$ set of all paths }} \langle \sigma \rangle$.
Therefore $x \in S_1(X) \Rightarrow x= \sum c_i \sigma_i$ and each $\sigma$ is infinite cyclic. What does $c_i \sigma_i$ mean, what does that constant operation on a path mean? And what does it mean for a path $\sigma$ to be infinite cyclic? And how do you define subtraction in $S_0(X)$ with $\sigma(1) - \sigma(0)$
I can see that the homomorphism would be something of $f : \sum_i c_i \sigma_i \mapsto \sum_i m_i x_i$, but to show $f(\sum_i(c_i+d_i)\sigma_i) = f(\sum_i c_1 \sigma_i) + f(\sum_i d_i \sigma_i)$, I feel that I would need more understanding of the bolded questions.
Please note that this question is related to the notions of homology on topological spaces. More precisely, you should look into singular homology if you want to have a good idea of why these computations are interesting. A good book for this is Algebraic Topology from Alan Hatcher.
Now, onto your question. Since I do not know what your background is, I am going to assume you have some basic knowledge in algebra and in point-set topology.
Let $X$ be a topological space. For an interval $I=\lbrack a,b\rbrack$, we say that $\gamma:I\to X$ is a path from $p\in X$ to $q\in X$ if $\gamma$ is continuous and $\gamma(a)=p$, $\gamma(b)=q$. We can always reparametrize our paths so that the interval of definition is $\lbrack 0,1\rbrack$. So, define $$\Omega^1(X)=\lbrace \gamma\; |\; \gamma\text{ is a path with interval }\lbrack 0,1\rbrack\rbrace$$ and form the free abelian group $S_1(X)$ generated by the set $\Omega^1(X)$. In other words, you only have formal sums of paths with coefficents in $\mathbb{Z}$ (I prefer to see it as a free $\mathbb{Z}$-module). Also define $S_0(X)$ to be the free abelian group generate by the points of $X$.
On a sidenote, for an alphabet $\mathscr{A}$, the free abelian group generated by $\mathscr{A}$ is defined as the set of formal finite $\mathbb{Z}$-linear combinations of elements of $\mathscr{A}$. You can also define it as $$\langle \mathscr{A}\rangle/\lbrack \langle \mathscr{A}\rangle,\langle \mathscr{A}\rangle\rbrack $$
i.e. the abeliannization of the free group $\langle \mathscr{A}\rangle$.
This definition is pretty abstract and let me explain why.
Unfortunalety, since you work with paths, this kind of definition does not carry much structure. A more meaningfull way is working with closed loops with a marked base point instead. You would then be defining the fundamental group of a topological space. I think if you are looking for a geometric intuition of what sums of path means, you would find it the there.
Let $$f:S_1(X)\to S_0(X)$$ be a map such that for any path $\gamma$, you have $f(\gamma)=\gamma(1)-\gamma(0)$. You want to define $f$ to be a homomorphism. I know that it feels "untrue from a geometric point of view" but since you have no intrisic structure on the path themselves, you need to ask $f(\gamma+\eta)=f(\gamma)+f(\eta)$. It then is a well defined homomorphism of $S_1(X)$ to $S_0(X)$ by definition.
Here is the moment where it is unevitable to realize that there is no "meaning" for these formal sums & map. This algebraic viewpoint is not tied to the topology of the space $X$ itself. The reason why the fundamental group is interesting is because "concatenation" is defined as the operation on two paths. Even when you proceed in singular homology, you do formal sums of simplices without any "adding of simplices". The geometric meaning is then carried by the borded operator. It transforms our previously defined sequence of abelian groups into a chain complex.
I hope my answer is satisfiying. I did my best to try to give indications on a what to do now. Sometimes with algebra, it is necessary to let go of the geometric notion of things and just accept definitions. For example, path are infinitely cyclic because they are defined this way.
And old professor of mine really insisted on the fact that it is impossible to understand definitions and that professionnal mathematicans are good at accepting them. Understanding why a definition is how it is of course legitimate and I encourage you to do so. As it is the case here, your definition needs to be sharpened to carry meaning. Further lectures are probably going to satisfy you.
I hope this helped.