Let $\theta : I \to \mathbb{R}^2$ be a regular plane curve with curvature $ |k_{\theta}|\leq1$ everywhere.
We now define a curve $\theta_{d}$ by stretching $\theta$ in one direction, i.e.,
$\theta = ( \theta_{1}(t), \theta_{2}(t))$,
$\theta_{d} = (2 \theta_{1}(t), \theta_{2}(t))$
How the curvature of $\theta_{d} , k_{d} $ relates to $\theta$ ?
Using the standard formula for curvature of a plane curve $\alpha(t)$, $$k = \frac{\|\alpha'(t)\times\alpha''(t)\|}{\|\alpha'(t)\|^3}\,,$$ we find that $$k_{\theta_d} = \frac{2k_\theta}{(4\theta_1'^2 + \theta_2'^2)^{3/2}}\,.$$ Not so informative, but since the speed of $\theta_d$ is between $1$ and $2$ and you gave the curvature bound on $\theta$, we have $$\frac{k_\theta}4 \le k_{\theta_d} \le 2{k_\theta}\le 2\,.$$