### Solution

#### Part $(1)$: Prove that $f\left(x\right)$ is an odd function
- The domain $\left(-1,1\right)$ is symmetric about the origin. Let's check if $f(-x) = -f(x)$:
$$f\left(-x\right) = \frac{-x}{1+(-x)^{2}} = -\frac{x}{1+{x}^{2}} = -f\left(x\right)$$
- Since $f\left(-x\right) = -f\left(x\right)$ holds true, we conclude that $f(x)$ is an odd function.

#### Part $(2)$: Determine and prove the monotonicity of the function $f\left(x\right)$ on $\left(-1,1\right)$
- Consider any two points $x_1$ and $x_2$ such that $-1 \lt x_{1} \lt x_{2} \lt 1$. We aim to compare $f(x_1)$ and $f(x_2)$:
$$f(x_{1}) - f(x_{2}) = \frac{x_{1}}{1+x_{1}^{2}} - \frac{x_{2}}{1+x_{2}^{2}} = \frac{(x_{1}-x_{2}) + x_{1}x_{2}(x_{2}-x_{1})}{(1+x_{1}^{2})(1+x_{2}^{2})}$$
$$= \frac{(x_{1}-x_{2})(1-x_{1}x_{2})}{(1+x_{1}^{2})(1+x_{2}^{2})}$$
- Since $x_1 0$. Thus:
$$(x_{1}-x_{2})(1-x_{1}x_{2}) < 0$$
- This implies $f(x_{1}) - f(x_{2}) < 0$, or equivalently, $f(x_{1}) < f(x_{2})$. Therefore, $f(x)$ is monotonically increasing on $(-1,1)$.

#### Part $(3)$: Solve the inequality $f\left(2x-1\right)+f\left(x\right) \lt 0$
- We have:
$$f\left(2x-1\right) < -f\left(x\right) = f\left(-x\right)$$
- This leads to the inequalities:
$$-1 < 2x-1 < -x < 1$$
- Solving for $x$ yields:
$$0 < x < \frac{1}{3}$$
- Thus, the solution set for the inequality is $\boxed{(0, \frac{1}{3})}$.