Could I understand the following results by interpreting curvature as acceleration? I saw these two theorems in the book "Riemannian Manifolds" by John M. Lee:
Thrm 1.5: Two smooth unit speed plane curves are congruent iff their curvatures are always the same.
Thrm 1.6: The total curvature of a closed curve is always 2π.
Could I interpret these two results by interpreting curvature as the signed magnitude of acceleration of the curve? (given by unit speed parameterisation)? In that case I could understand them by the following mechanical reasoning:
Thrm 1.5: Two curves are the same iff they have the same acceleration at all points, which seems natural.
Thrm 1.6: The total curvature, by definition, is the integral of curvature all along the curve, which under my interpretation could be thought as the integral of acceleration of a "pencil tip" over a given period of time, such that it traces out the curve and ends at the given starting point. Since the integral of acceleration is simple velocity, I could interpret Thrm 1.6 as saying the pencil tip's velocity vector has rotated by 2π when drawing the curve, which seems natural. When you finished drawing the curve, your pencil tip has done all crazy stuffs and ended at the same direction as you began. So it seems natural that the answer should be a multiple of 2π (in this case exactly 2π).
Also, under this understanding, could I understand why it has to be 2π in Thrm 1.6 instead of 4π or 6π for some crazy curves that have sharp vertices, where the pencil tip rotates drastically?
Any help will be appreciated! It would be even better if you could provide an explanation in simple language as I lack even the knowledge of Metric Spaces.
 A: 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.
