To solve this problem, let's break it down step by step:

1. Define Set A: Given $A=\{y|y=\log _{2}x,x > 4\}$, we need to understand the range of $y$. Since $\log_{2}x$ is an increasing function, and $x > 4$, we can substitute $x = 4$ into the logarithm to find the lower bound of $y$. This gives us $y = \log_{2}4 = 2$. Therefore, the values of $y$ are greater than $2$, which means $A=\{y|y > 2\}$.

2. Define Set B: Given $B=\{x|x^{2}-3x+2 < 0\}$, we factor the quadratic equation to $(x-1)(x-2) < 0$. This inequality holds true for $x$ in the interval $(1, 2)$, meaning $B=\{x|1 < x < 2\}$.

3. **Complement of Set A in $\mathbb{R}$**: The complement of $A$ in the real numbers, denoted as $\complement_{R}A$, includes all real numbers not in $A$. Since $A$ contains numbers greater than $2$, its complement will contain numbers less than or equal to $2$, which is $\complement_{R}A=\{y|y\leqslant 2\}$.

4. **Union of $\complement_{R}A$ and B**: The union of sets $\complement_{R}A$ and $B$ combines the elements of both sets. Since $\complement_{R}A$ includes all numbers up to $2$ and $B$ includes numbers between $1$ and $2$, their union covers all numbers up to $2$, including $2$ itself. Therefore, $(\complement_{R}A)\cup B=(-\infty ,2]$.

Thus, the correct answer, following the rules and formatting, is $\boxed{C}$.