Could I understand the following results by interpreting curvature as acceleration?

Solution 1:

First, you missed some important hypotheses when you quoted these theorems.

Thrm 1.6: The total curvature of a closed curve is always 2π.

What Theorem 1.6 actually says is that if $\gamma:[a,b]\to\mathbb R^2$ is a unit-speed simple closed curve such that $\boldsymbol{\gamma'(a)=\gamma'(b)}$ and the curvature is computed with respect to the inward-pointing normal, then the total curvature is $2\pi$. Without these additional hypotheses, the result is false.

Thrm 1.5: Two smooth unit speed plane curves are congruent iff their curvatures are always the same.

What Theorem 1.5 says, more accurately, is that if $\gamma$ and $\tilde\gamma$ are smooth, unit-speed plane curves with the same parameter interval and chosen unit normals $N$ and $\widetilde N$, respectively, then $\gamma$ and $\tilde \gamma$ are congruent by a direction-preserving congruence if and only if the signed curvatures $\kappa_N$ and $\tilde \kappa_{\widetilde N}$ at corresponding parameter values are equal.

With these corrections, your intuition is correct: For a unit-speed plane curve, the signed curvature at $t$ is exactly the $t$-derivative of the angle $\theta(t)$ that $\gamma'(t)$ makes with the positive $x$-axis. At the top of page 275 in my book, I prove the following formula for plane curve with respect to an arbitrary Riemannian metric: $$ \kappa_N = \theta' - \omega(\gamma'), $$ where $\omega$ is the "connection $1$-form," which expresses the Levi-Civita connection of the given Riemannian metric. In the special case when the metric is the Euclidean metric on $\mathbb R^2$, this form is identically zero, so the signed curvature is exactly the derivative of the angle function.