I find this a natural question. Particularly because I recently heard that I would be teaching group theory for advanced undergrads next year (or the year after), so I want to test my motivational skills here :-)
Indeed, one of the aspects of representation theory is to study the groups being represented. It is nice that we can take some abstract group, and replace its elements with objects that are easier to compute with: matrices (as in linear representations) or permutations (as in groups acting on sets). To that end it is helpful (if not imperative) that the representation is faithful lest we draw mistaken conclusions about identities of elements of the said group. Even if the action is not faithful we can slice off the kernel of the action as a part of a divide-and-conquer approach to understanding the group better, if/when so inclined.
On the other hand we also want to understand what kind of objects can be acted upon by a given group (either via permutations or linear transformations). There the focus is different. If we look at objects being permuted by a group, the first divide-and-conquer step is to study the objects one orbit at a time, as the orbits don't interact (but e.g. their sizes are constrained by the size of the acting group). If we look at vector spaces acted upon by a group of linear transformations, then the divide-and-conquer approach tells us to study subspaces stable under the group, and check whether we can build the whole space from atomic subspaces (often we can).
So representation theory is in some sense a two-way street: extract information about the group by studying its possible actions, and OTOH extract information about a collection of objects given that they are acted upon by a given group in some meaningful way.
Possibly the best known result in the former direction is the Burnside's $p^aq^b$-theorem telling us that a finite group whose order has only two prime divisors is necessarily solvable. This is likely covered in all texts. Going the other way let me lead off with a simple example. Let $V$ be the space of complex functions on the real line with period $2\pi$. If $f(t)$ is such a function, then so is the function $g\cdot f$ defined by 180 degree phase shift: $(g\cdot f)(t)=f(t+\pi)$. As the functions in $V$ have period $2\pi$, we see that $g\cdot(g\cdot f)=f$ for all $f$ meaning that the cyclic group $C_2=\langle g\rangle$ of order two generated by $g$ acts on $V$. What can we say about this action? Representation theory tells us that there are the following kind of special functions in relation to this action. The space
$$
V^+=\{f\in V\mid g\cdot f=f\}
$$
of functions that have $\pi$ as a period, and the space
$$
V^-=\{f\in V\mid g\cdot f=-f\},
$$
i.e. the function $f$ with the property $f(t+\pi)=-f(t)$ for all $t$. Representation theory also tells us that $V=V^+\oplus V^-$. It is an easy exercise that given any function $f\in V$, then the function
$$
f^+(t)=\frac{f(t)+f(t+\pi)}2\in V^+,
$$
and
$$
f^-(t)=\frac{f(t)-f(t+\pi)}2\in V^-.
$$
Furthermore $f=f^++f^-$. Once you reach the idempotents of the group ring in your studies, this should give you a deja vu -experience :-) At that point an easy exercise for you is to figure what happens, if $g$ acts by a phase shift by $2\pi/n$ for some natural number $n>2$.
A more high-browed example of this latter direction comes from particle physics. There it is imperative that the symmetries of the nature are not broken. Quantum physics uses vectors in some space to represent elementary particles. What then happens is that particles can only react in such a way that the tensor product of those vectors has a non-trivial component acted upon trivially by the group. IOW: "trivial representation = invariants" is a very important representation - fully agree with Mariano.
Caveat: I did go on a hyperbole with the last example, and the painted picture may not be quite accurate. Anyway, they were handing out Nobel prizes in physics for that stuff in the 60s and 70s :-)
The $W_i$ are just the eigenspaces of the matrix $\rho(g)$, with eigenvalue $\omega^i$ on $W_i$. In this case, there is a neat way to explicitly write down the eigenspaces: the vector $$v_i=(1,\omega^i,\omega^{2i},\dots,\omega^{(n-1)i})$$ is easily seen to be an eigenvector for $\rho(g)$ with eigenvalue $\omega^i$. So, $W_i$ is just the span of $v_i$ for each $i$.
(In case the formula for $v_i$ seems miraculous, note that it is easy to derive: if $v=(a_0,a_1,\dots,v_{n-1})$ then $\rho(g)v=(a_1,a_2,\dots,a_0)$ so to be an eigenvector with eigenvalue $\omega^i$ we must have $a_1=\omega^ia_0$, $a_2=\omega^ia_1=\omega^{2i}a_0$, and so on.)
Best Answer
The "representation on the regular representation" is probably just a typo for "the regular representation". This is the representation on the vector space $\mathbb CG$ in which each element of $G$ acts by multiplication from the left (on the standard basis vectors of $\mathbb CG$). That representation is faithful because every non-identity element of $G$ moves some vectors in $\mathbb CG$, for example any element of $G$ (considered as a basis vector in $\mathbb CG$). Finally, if we decompose the regular representation into irreducible summands, then each non-identity element must act nontrivially in at least one of those irreducible representations.