In the book of E. Bloch, at page 171, it is stated that
Corollary:
Let $U\subseteq \mathbb{R}^2$ be an open subset, and let $f: U \to \mathbb{R}^3$ be a smooth map. If for $p \in U$, $Df(p)$ has rank 2, then there is an open subset $W$ of $U$ s.t $f|_W$ is injective, and $Df(q)$ for all $q\in W$ has rank 2.
However, when we define a coordinate patch in page 203,
Definition:
Let $U\subseteq \mathbb{R}^2$ be an open set. A smooth map $x: U\to \mathbb{R}^3$ is a coordinate path if it is injective and $x_1 \times x_2 \not = 0$ for all $p\in U$.
Equivalently to the above conditions are that $x_1 $ and $x_2$ are linearly independent, or that $Dx$ has rank 2 for all $q\in U$.
But by the corollary, if $x$ has rank 2, by choosing a "smaller" neighbourhood of $p$, we can make it an injective map, so isn't the condition for $x$ to be injective redundant in here ? or Am I missing something ?
It isn't redundant because being locally injective doesn't mean that you are injective. If you start with a map $x \colon U \rightarrow \mathbb{R}^3$ defined on $U$ such that $Dx(p)$ has rank two for any $p \in U$, then, give a $p \in U$ you can find a neighborhood $W$ of $p$ such that $x|_{W} \colon W \rightarrow \mathbb{R}^3$ is injective. It doesn't mean that $x$ itself is injective as it is possible to have $q' \in W, q'' \in U$ such that $x(q') = x(q'')$. The point $q''$ can't also be in $W$ because $x|_{W}$ is injective but nothing forbids the existence of a point in $U \setminus W$ which has the same image as a point in $W$.
For a concrete example, consider $x(\theta,z) = (\cos \theta, \sin \theta, z)$ defined on $\mathbb{R}^2$, whose image is an infinite cylinder. The differential has rank two anywhere, but it is only locally injective and not injective. If you drop the condition of being injective, you are left with a useful notion in differential geometry which is called immersion but it is not the same as being a coordinate patch.