To analyze the function $y=x^{2}-\log _{a}(x+1)-4x+4$ and determine the range of real numbers $a$ for which the function values are all less than $0$ when $x\in \left(1,2\right)$, we proceed in a step-by-step manner.

1. **Considering the case when $0 0$ because the logarithm of a positive number to a base between $0$ and $1$ is negative, and the negative of a negative number is positive.
- Next, we analyze $x^{2}-4x+4$. Completing the square gives $x^{2}-4x+4=(x-2)^{2}$. Since a square is always non-negative, $(x-2)^{2} \geq 0$. Thus, for any $x$, $x^{2}-4x+4 > 0$ as long as $x \neq 2$.
- Combining these results, we find that $y = (x-2)^{2} - \log_{a}(x+1) > 0$ for $0 1$:**

- When $a > 1$, to analyze the behavior of $y=x^{2}-\log _{a}(x+1)-4x+4$ in the interval $(1,2)$, we consider the monotonicity of its components:
- The term $x^{2}-4x+4$ simplifies to $(x-2)^{2}$, which is monotonically increasing for $x > 2$ and decreasing for $x 1$ because as $x$ increases, $\log _{a}(x+1)$ increases and its negative decreases.
- Combining these observations, we deduce that $y$ is monotonically decreasing on $(1,2)$.
- For the function values to be all less than $0$ in $(1,2)$, it suffices to ensure that the maximum value of $y$ in this interval does not exceed $0$. The maximum occurs at the left endpoint $x=1$ due to the monotonic decrease:
- At $x=1$, $y_{\max}=y|_{x=1}=1^{2}-\log _{a}(2)-4(1)+4=1-\log _{a}(2)$.
- Requiring that $y_{\max} \leq 0$ leads to $1-\log _{a}(2) \leq 0$, or $\log _{a}(2) \geq 1$.

3. **Solving for $a$:**

- The inequality $\log _{a}(2) \geq 1$ can be rewritten as $a^{1} \geq 2$, giving $a \geq 2$.
- However, considering the overall conditions and the behavior of the function, we refine this to $1 1$, we conclude that the range of $a$ for which the function $y=x^{2}-\log _{a}(x+1)-4x+4$ has all values less than $0$ for $x\in (1,2)$ is $\boxed{(1,2]}$.