Yes, there is a set of definitions that can bring consistency to the terminology.
A curve $\gamma\colon I \to \mathbb{R}^n$ is smooth iff it is $C^\infty$ (or $C^p$ for some authors).
A curve $\gamma\colon I \to \mathbb{R}^n$ is an immersion iff it is $C^\infty$ and $\gamma'(t) \neq 0$.
A curve $\gamma\colon I \to \mathbb{R}^n$ is simple iff $\gamma$ is injective on the interior of $I$.
(The reason for requiring injectivity on the interior, rather than on the whole interval, is so that we may legitimately speak of "simple closed curves." That is a simple closed curve $\gamma\colon [a,b] \to \mathbb{R}^n$ is a map that is injective on $[a,b)$ and has $\gamma(a) = \gamma(b)$.)
Granted, these definitions aren't 100% universal, but I think they're fairly standard. The term "regular" is sometimes used as a synonym for "immersion," but then again, "regular" can mean other things in other contexts, too (which is why I prefer to avoid the term altogether if I can help it).
And as you point out, some authors refer to a curve as the image of such a map, rather than the map itself.
This is exercise $2.2.7$ in Pressley's Elementary Differential Geometry (which has solution sketches in the back of the text). I've expanded a bit on the solution provided.
Hints:
For signed curvature, what is the definition of signed curvature in terms of your signed normal and tangent? Remember that you're trying to find $\kappa_{s_{\epsilon}}$.
For the normal line part, what does it mean for a line to be tangent to $\epsilon$? What does a normal line at a specific point $s_0$ look like? We calculated the tangent to $\epsilon$ in the first part.
(More complete answer below)
$$\epsilon (s)=\gamma (s)+\frac{1}{\kappa_s(s)}n_s(s)$$
$\textbf{1.}$ We show that the arc length of $\epsilon$ is $\frac{-1}{\kappa_s(s)}$ up to a constant. First off, I'm going to forget about the $s$ notation-wise so assume everything is a function of $s$ unless stated otherwise.
Well, we take the derivative of $\epsilon$ with respect to $s$ to get the arc length:
$$\dot{\epsilon} = \dot{\gamma} + \frac{1}{\kappa_s}(-\kappa_s \mathbf{t})-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}=-\frac{\dot{\kappa_s} \mathbf{n_s}}{\kappa_s^2}$$
(recall $\mathbf{\dot{t}} = \kappa_s \mathbf{n_s} \implies \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$ for a unit-speed curve since $\mathbf{n_s} \cdot \mathbf{t} = 0$ so $\mathbf{\dot{n_s}} \cdot \mathbf{t} + \mathbf{n_s} \cdot \mathbf{\dot{t}} = 0 \implies \mathbf{\dot{n_s}} \cdot \mathbf{t}= - \kappa_s(\mathbf{n_s \cdot n_s}) = -\kappa_s \implies \mathbf{\dot{n_s}}(\mathbf{t}\cdot \mathbf{t}) = \mathbf{\dot{n_s}} = -\kappa_s \mathbf{t}$)
Now, the arc length is given by
$$u=\int \| \dot{\epsilon} \| \,ds = \int \frac{\dot{\kappa_s}}{\kappa_s^2} \,ds = -\frac{1}{\kappa_s} + C$$
Note that the second equality holds since we assumed $\dot{\kappa_s} > 0$.
$\textbf{2.}$ We calculate the signed curvature $\kappa_{s_{\epsilon}}.$ Recall the signed curvature is the rate at which the tangent vector rotates. In particular,
$$\mathbf{\dot{t}}_{\epsilon} = \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$ In this case, we take the tangent vector to be $\mathbf{t}_{\epsilon}=-\mathbf{n_s}$. Rotating the tangent vector counterclockwise by $-\pi/2$ gives us our signed unit normal. In particular, the signed normal is just $\mathbf{n_s}_{\epsilon}=\mathbf{t}$. Now,
$$\frac{d (-\mathbf{n_s})}{\,du} = \kappa_s \mathbf{t} \frac{ds}{du} = \kappa_s \mathbf{t} \frac{\kappa_s^2}{\dot{\kappa_s}}= \frac{\kappa_s^3}{\dot{\kappa_s}}\mathbf{t}= \kappa_{s_{\epsilon}}\mathbf{n_s}_{\epsilon}$$
In particular, since the derivative of the tangent vector is the signed curvature times the signed unit normal, dotting the derivative of the tangent vector with the signed unit normal gives the result. That is, take the dot product of the above expression with $\mathbf{t}$ to get the signed curvature of $\epsilon$:
$$\frac{\kappa_s^3}{\dot{\kappa_s}}$$
$\textbf{3.}$ We show that all normal lines to $\gamma$ are tangent to $\epsilon$.
Well, let's look at a point on the normal line at $\gamma(s_0)$ for some arbitrary $s_0$. It looks like $\gamma(s_0) + C\mathbf{n_s}(s_0)$ for some $C$. Since $\epsilon(s_0) = \gamma(s_0) + \frac{1}{\kappa_s(s_0)}\mathbf{n_s}(s_0)$, the intersection occurs when $C=\frac{1}{\kappa_s(s_0)}$. Well, we calculated the tangent of $\epsilon$ at $s_0$:
$$\dot{\epsilon}(s_0)=-\frac{\dot{\kappa_s(s_0)} \mathbf{n_s}(s_0)}{\kappa_s^2(s_0)}$$
so that the tangent there is parallel to $\mathbf{n_s}(s_0)$.
$\textbf{4.}$ Regarding the evolute of the cycloid, this is just a computation, with a lot of the steps highlighted above. Regarding the reparameterization, consider $\tilde{t} = t + \pi$.
Best Answer
EDITED
These definitions could be written far more clearly. So your first definition (definition 0?) should say that $\beta$ is an involute of $\gamma$ if for each $s$, $$\beta(s) = \gamma(s) + \lambda(s)T_\gamma(s) \qquad\text{and}\qquad T_\beta(s)\cdot T_\gamma(s)=0.$$ The first condition says $\beta$ lies on the tangent line of $\gamma$, and the second says their tangent vectors are orthogonal. (Note that I will use $s$ as an arclength parameter on $\gamma$, but it is far from one for $\beta$.) This is now the definition as I gave it in the comments.
Reversing letters so as to be less confusing, definition 1/2 says that $\gamma$ is an evolute of $\beta$ (i.e., $\beta$ is an involute of $\gamma$) if the principal normal lines of $\beta$ are tangent to $\gamma$, i.e., if $$\gamma(s) = \beta(s) + \mu(s)N_\beta(s) \qquad\text{and}\qquad N_\beta(s) = \pm T_\gamma(s).$$
A standard exercise is this: From definition 0, we deduce that $\beta$ is an involute of $\gamma$ if and only if $\beta(s)=\gamma(s)+(c-s)T_\gamma(s)$ for some constant $c$. Indeed, it is immediate from the Frenet equations that $T_\beta$ is $\pm N_\gamma$. Interchanging, by definition 2, $\beta$ is an evolute of $\gamma$ if and only if $T_\gamma = \pm N_\beta$. This gives you definition 1.
The derivation in the linked post is correct. There's no problem when $\tau=0$; you just take $c=\pi/2$. The careful statement — details matter! — is that there is some value of the constant $c$ for which the curve will be an evolute, not all.