I'm interpreting the statement as follows:
There is a finite collection of distinguished coordinate charts $U_i$ covering $M$ where each $U_i\cup U_j$ is contained in another (perhaps nondistinguished) coordinate chart $W_{ij}$.
I am not necessarily claiming that $W_{ij}\cup W_{kl}$ is contained in a chart $Z_{ijkl}$ and each $Z_{ijkl}\cup Z_{mnop}$ is contained in a chart ...., just that the union of the $U$s are contained in charts.
Here's a proof that such things always exist, though it may use tools you don't have access to.
Equip $M$ with any Riemannian metric $g$. This gives rise to several functions. First, for each point, there is a map called the exponential map, $\operatorname{exp}_p :T_p M\rightarrow M$ with the property that $\operatorname{exp}_p$ is a diffeomorphism onto its image when restricted to a small enough ball around the origin in $T_p M$. It is a fact that when $\operatorname{exp}_p$ is restricted to a ball of radius $r$ in $T_p M$, the image consists of all points in $M$ a distance $r$ away from $p$.
Another important function coming from a choice of metric is the function $\operatorname{inj}:M\rightarrow\mathbb{R}$. It is called the injectivity radius and defined as the largest radius of a ball around the origin in $T_p M$ such that $\operatorname{exp}_p:B(r)\rightarrow M$ is a diffeomorphism onto its image. It is known that $\operatorname{inj}$ is a continuous function and, on a compact manifold, bounded away from $0$.
Let $\rho$ denote the minimum value of $\operatorname{inj}$ on $M$. Then, by definition, at every point $p$, if we restrict $\operatorname{exp}_p$ to the ball of radius $\rho$, it's a diffeomorphism onto its image. In particular, we can use $\operatorname{exp}_p(B(\rho))$ as a chart on $M$.
Let $U_p$ denote $\operatorname{exp}_p(B(\frac{\rho}{2}))$. I claim that the collection of $\{U_p\}$ has the property that the union of any 2 are contained in a coordinate chart and that they cover $M$. Once we establish all this, use compactness of $M$ to extract a finite subcollection of the $U_p$, giving the desired collection of charts.
First, $p\in U_p$ since $\operatorname{exp}_p(0) = p$ always. Thus, these do cover.
Now, why do they have the property that the union is contained in a coordinate chart? We'll break into 2 cases depending on whether or not $U_p\cap U_q$ is empty or not.
If $U_p\cap U_q =\emptyset$, then $U_p\cup U_q$ is a (disconnected) chart. The diffeomorphism between it and an open subset of $\mathbb{R}^n$ is given by $\begin{cases} \operatorname{exp}_p^{-1}(r) & r\in U_p \\ \operatorname{exp}_q^{-1}(r) + v & r\in U_q\end{cases}, $ where $v$ is some vector of very large (compared to $\rho$) length. The point of $v$ is to shift the ball $\operatorname{exp}_q^{-1}(U_q)$ in $T_q M\cong \mathbb{R}^n$ far enough away from the origin so that it wont intersect the image of $\operatorname{exp}_p^{-1}(U_p)$ in $T_p M \cong\mathbb{R}^n$
On the other hand, if $U_p\cap U_q \neq \emptyset$, choose $r\in U_p\cap U_q$. I claim that $U_p\cup U_q \subseteq \operatorname{exp}_r(B(\rho))$. To see this, we'll use the triangle inequality. Using $d$ to denote distance, we have for any $s\in U_p$, that $d(s,r)\leq d(s,p) + d(p,r) < \frac{\rho}{2} + \frac{\rho}{2} = \rho$. This same argument works for $U_q$, so we have $U_p\cup U_q\subseteq \operatorname{exp}_r(B(\rho))$.