Let finitely many charts of a manifold $M$ be given by
$\displaystyle\Phi_j \colon T_j\rightarrow V_j\subseteq M , \ j=1,...,m$
where $\bigcup_{j=1}^m V_j=M$. We then choose a partition of unity satisfying
$\displaystyle \alpha_j\colon M\rightarrow\mathbb{R},\ j=1,...,m\\ 0\le\alpha_j\le 1, \alpha_j|_{M\backslash V_j}=0$
and the additional usual properties of partitions of unity. The integral of $f\colon M\rightarrow \mathbb{R}$ over $M$ is defined by
$\displaystyle \int_Mf(x)\mathrm{d}S(x)= \sum_{j=1}^m \int_M \alpha_j(x)f(x)\mathrm{d}S(x)=\sum_{j=1}^m\int_T(\alpha_jf)(\Phi_j(t))\sqrt{g_j(t)}\mathrm{d}t$
where $g_j$ is the gram determinant of the respective chart.
I have two questions. Firstly, the charts are merely requiered to cover $M$, but what if the $\Phi_j(T_j)$ are not disjoint, wouldn't this mean that some parts of the manifold are integrated over more that once?
Secondly, why is it necessary to use a partition of unity? I might aswell just define
$\displaystyle \int_Mf(x)\mathrm{d}S(x)=\sum_{j=1}^m\int_Tf(\Phi_j(t))\sqrt{g_j(t)}\mathrm{d}t$.
I am not sure why we need to force that $f$ is zero outside of the charted area for a chart. It does not matter what it is outside of the charted area, since it is never evaluated there.
The idea is that you have $\displaystyle f=\sum_{j=1}^m \alpha_jf$, so it makes sense to define the integral of $f$ as the sum of integrals on $f_j:=\alpha_jf$. So you don't have to worry about $\Phi_j(T_j)$ not being disjoint because each term in your sum is an integral on a different function.
As for the second question, note that the expression $\int_Tf(\Phi(t))\sqrt{g_j(t)}\mathrm{d}t$ can only be defined if the support of $f$ is contained in $\Phi(T)$. You need a chart $\Phi$ that satisfies that. But sometimes a chart like that does not exist. That's why we need a partition of unity, that way you write $f$ as a sum of functions $f_j$ that satisfy $supp(f_j) \subset \Phi_j(T_j)$ for each $j$.