Partition of unity and Riemannian metric

In the proof of existence of Riemannian metric on a smooth manifold M I stumbled upon this:

M is smooth => M has a smooth atlas {${U_i,\phi_i}$}. Now we define partition of unity: It is a set of functions {$f_{i} :M\rightarrow[0,1]$} so that $supp(f_i)\subset U_i$. Now it is $$\sum_i f_i(x)=1$$ for all $x\in M$.

My question is, why does $\sum_i f_i(x)=1$? What if there is a point p$\in supp(f_i)\bigcap supp(f_j)$for $j\neq i$, and let's say $f_i(p)=1=f_j(p)$ and $0=f_k(p)$ for $k\neq i\neq j$. Then $\sum_i f_i(p)=2$.

Or do we define a partition of unity so that $\sum_i f_i(x)=1$ holds for all x?

As you indicate in your last question, we define a partition of unity to obey $\sum f_i (x) = 1 ~\forall x \in M$. This is almost a verbatim translation of "partition of unity":
For an example, let $M=(0,1)$. One possible partition of unity would be $f_1(x) = 1-x$ and $f_2(x) = x$. Note that this would require that $U_1 = U_2 = (0,1)$ in order for $supp(f_i) \subset U_i$ to hold. This is also somewhat of a lie, as we usually also require the support of a function to be compact.