$\newcommand{\Reals}{\mathbf{R}}$Let $\Gamma_{0}$ denote $\Gamma$ with the point $\gamma(a) = \gamma(b) = \beta(c) = \beta(d)$ removed.
Each of $\gamma:(a, b) \to \Gamma_{0}$ and $\beta:(c, d) \to \Gamma_{0}$ is a diffeomorphism, so $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is a diffeomorphism (as smooth as the less smooth of the two paths),[*] and extends to the endpoints uniquely by continuity. The extended overlap map is a reparametrization.
For the starred point, if you need to go lower-level than this the important point is that $\gamma_{*}:T(a, b) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point because $\gamma$ is regular, and similarly $\beta_{*}:T(c, d) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point. Consequently, $( \beta^{-1}\circ\gamma)_{*} = (\beta^{-1})_{*}\circ\gamma_{*}:T(a, b) \to T(c, d)$ is an isomorphism of one-dimensional spaces at each point.
Added: We know the function $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is bijective and satisfies $\gamma = \beta \circ (\beta^{-1} \circ \gamma)$. As indicated by the comments, we need to show $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.
Proposition: If $\gamma(t_{0}) = \beta(\tau_{0})$ for some $t_{0}$ in $(a, b)$ and $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$ in $(c, d)$, then $\gamma'(t_{0})$ and $\beta'(\tau_{0})$ are proportional, i.e., each is a (non-zero) multiple of the other.
Proof: Pick an orthonormal basis $(e_{j})_{j=1}^{n-1}$ for the hyperplane orthogonal to $\gamma'(t_{0})$ and define the smooth mapping $F:\Reals^{n} \to \Reals^{n}$ by
$$
F(u_{1}, \dots, u_{n-1}, t) = u_{1}e_{1} + \cdots + u_{n-1}e_{n-1} + \gamma(t).
$$
Since $Df = [e_{1}\ \cdots\ e_{n-1}\ \gamma'(t)]$ is non-singular at $T_{0} := (0, \dots, 0, t_{0})$, the inverse function theorem guarantees there is a neighborhood $U$ of $T_{0}$ in which $F$ is smoothly invertible. Write $V = F(U)$. By construction, the functions $\phi_{j} := u_{j} \circ F^{-1}$ are smooth, vanish simultaneously on the trace of $\gamma$ and nowhere else, and have linearly independent gradients in $V$. That is, the trace of $\gamma$ intersected with $V$ is precisely the zero set of $\Phi := (\phi_{1}, \dots, \phi_{n-1})$, and the kernel of this mapping, $\ker D\Phi(T_{0})$, is one-dimensional. Since $\beta$ has the same trace as $\gamma$, the non-zero vector $\beta'(\tau_{0})$ lies in this same one-dimensional subspace. That is, $\beta'(\tau_{0})$ and $\gamma'(t_{0})$ are proportional. This completes the proof of the proposition.
Write $\gamma = (\gamma_{k})_{k=1}^{n}$ and $\beta = (\beta_{k})_{k=1}^{n}$, and note that $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$ for each $k$. Let $t_{0}$ be an arbitrary point of $(a, b)$ and put $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$. Because $\beta$ is regular, there exists an index $k$ such that $\beta_{k}'(\tau_{0}) \neq 0$. The preceding proposition guarantees $\gamma_{k}'(t_{0}) \neq 0$. By the one-variable inverse function theorem, there exists a $\delta > 0$ such that $\beta_{k}$ is smoothly invertible in $(\tau_{0} - \delta, \tau_{0} + \delta)$. Consequently, $\beta_{k}^{-1} \circ \gamma_{k}$ is smoothly invertible in some neighborhood of $t_{0}$ as a composition of smoothly invertible functions. But $\beta^{-1} \circ \gamma = \beta_{k}^{-1} \circ \gamma_{k}$ (!) in this neighborhood by applying $\beta_{k}^{-1}$ to both sides of $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$. Since $t_{0}$ was arbitrary, $\beta^{-1} \circ \gamma$ is locally smoothly invertible at each point.
In particular, the preceding paragraph shows that the bijection $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is continuous. A continuous bijection whose domain is an interval is strictly monotone (by the intermediate value theorem), and a strictly monotone function between bounded intervals extends uniquely by continuity to the endpoints. In summary, we have shown that $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.
To (hopefully) clarify notation, for a map $F:\mathbb{R}^k\to\mathbb{R}^l$, let $D_xF$ denote the derivative of $F$ at $x$, in the sense of multivariable calculus.
Let $\gamma_1,\gamma_2$ be smooth curves with $\gamma_1(0)=\gamma_2(0)=p$, suppose that $D_0(\varphi\circ\gamma_1)=D_0(\varphi\circ\gamma_2)$ for some chart $\varphi$, and let $\psi$ be any other chart containing $p$. Since $M$ is a smooth manifold, the transition function $\tau=\psi\circ\varphi^{-1}$ is a deffeomorphism between open subsets of $\mathbb{R}^n$. We can compute the coordinate derivative of $\gamma_1$ in the new chart.
\begin{align*}
D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\varphi^{-1}\circ\varphi\circ\gamma_1) \\
=&D_0(\tau\circ(\varphi\circ\gamma_1))
\end{align*}
Applying the chain rule:
\begin{align*}
=&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1)
\end{align*}
We see that $D_0(\psi\circ\gamma_1)$ is related to $D_0(\varphi\circ\gamma_1)$ by the linear map $D_{\varphi(0)}\tau$. The same is true of $\gamma_2$. As a result, of the derivatives are equal in one chart, they are equal in any other:
\begin{align*}
D_0(\varphi\circ\gamma_1)=&D_0(\varphi\circ\gamma_2) \\
\implies D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1)=&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_2) \\
\implies D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\gamma_2) \\
\end{align*}
Best Answer
As you wrote, the image of both is the unit circle. Or you could note that $\gamma_2(t) = \gamma_1(2t)$ for $t \in [0,\pi]$ and $\gamma_1(2t-2\pi)$ for $t \in [\pi,2\pi]$.
$\gamma_1$ is one-to-one, and so would be $\gamma_1 \circ \phi$. $\gamma_2$ is not.