One of the definition of simplicial approximation says that: a simplicial map $h:|K|\rightarrow|L|$ is a simplicial approximation of a continuous map $f:|K|\rightarrow|L|$ if and only if $$\forall x\in|K|, h(x)\in Car_{L}f(x),$$ where $Car_{L}f(x)$ means the carrier simplex of $f(x)$.
My question is that does this mean that if $f(x)$ is a vertex implies x is also a vertex?
Since it seams that if $f(x)$ is a vertex of $L$, then $h(x)\in Car_{L}f(x)=f(x)\ $ which implies $h(x)=f(x)\ $?
Then if the preimage of a vertex of $L$ under f is at some irrational position of $K$, how is the bisection progress guarantee the existence of a simplicial approximation?
The definition you gave seems to be equivalent to the following $$ f(|\mathring\sigma|)\subseteq\text{st}(h(\sigma))\ \text{ for all } \sigma\in K $$ For if $h$ and $f$ satisfy the property given in your question, then if $x\in|\mathring\sigma|$, we have $h(x)\in\text{Car}(f(x))$, and that means $h(\sigma)$ is a face of the simplex whose interior contains $f(x)$, so $f(x)$ is in the open star of $h(\sigma)$. On the other hand, if the above formula holds for every $\sigma\in K$, then given an $x\in|\mathring\sigma|$, it is $f(x)$ in the interior of some simplex with face $h(\sigma)$, and thus $h(x)$ is in the carrier of $f(x)$.
The map $f$ need not map vertices to vertices to have a simplicial approximation. Maybe the simplest example is where $K=\{\{v\}\}$ and $L=\{\{v\},\{w\},\{v,w\}\}$ and $f:|K|\to|L|$ is the map sending $v$ to the point in the middle of the edge $|\{v,w\}|$. Then the inclusion $K\hookrightarrow L$ is a simplicial approximation of $f$.
Regarding your edited question: If $f(x)$ is a vertex, then as you correctly noticed $\text{Car}_Lf(x)=f(x)=h(x)$. When $h$ is injective, then this implies that $x$ is a vertex. If $h$ is not injective, then $x$ need not be a vertex. For example $h:L\to K$ (the complexes defined as above), is a simplicial map sending every vertex to $v$, so it also sends $\{v,w\}$ to $\{v\}$, and then $f=|h|$ sends $\frac12v+\frac12w$ to $v$