So, why do some proofs change the boundary and some do not? I mean how are these equivalent?
They aren't: Common proofs of 'Noether's theorem' often only consider certain limits of it. There's also a tendency to hide intricacies behind notation.
In the following, I'll present an elementary proof of a simple version of Noether's first theorem in 1 dimension in a way that should generalize to what's called the the field-theoretic version on Wikipedia by going from $t$ to $x^\mu$ and $q$ to $\varphi^A$.
The Lagrangian will be a function
$$
L = L(x,v,t)
$$
and the action a functional
$$
S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt
$$
Proposition. If the transformation
$$ t\to t'(t) = t + \epsilon T(t) $$
$$ x\to x'(x,t) = x + \epsilon X(t)$$
$$
q'(t') = q(t(t')) + \epsilon X(t(t'))
$$
is a quasi-symmetry of the action
$$
\delta S \approx \Delta K
$$
on-shell (ie assuming the equations of motion), then there is a conserved quantity
$$
\frac{d}{dt} \left( \frac{\partial L}{\partial v} (X - \dot q T) + LT - K \right) \approx 0
$$
Here,
$$ \delta S = \frac{d}{d\epsilon}\Big|_{\epsilon=0} S[q'] $$
$$ \Delta K = K(q(t_2),t_2)-K(q(t_1),t_1) = \int_{t_1}^{t_2} \frac{dK}{dt} dt $$
Proof.
\begin{align}
\delta S &= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t'(t_1)}^{t'(t_2)} L(q'(t'),\frac{d}{dt'}q'(t'),t')\,dt'
\\&= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t_1}^{t_2} L(q(t) + \epsilon X(t),\left( \frac{dt'}{dt} \right)^{-1}\frac{d}{dt}(q(t) + \epsilon X(t)),t + \epsilon T(t)) \,dt'(t)
\\&= \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v}\frac{d}{d\epsilon}\Big|_{\epsilon=0}\frac{\dot q + \epsilon \dot X}{1 + \epsilon \dot T} + \frac{\partial L}{\partial t} T \right)\,dt + L \frac{d}{d\epsilon}\Big|_{\epsilon=0}d(t + \epsilon T) \right]
\\&= \int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v} (\dot X - \dot q \dot T) + \frac{\partial L}{\partial t} T + L \dot T\right]\,dt
\end{align}
Using
$$
\frac{\partial L}{\partial v} \dot X = \frac{d}{dt}\left( \frac{\partial L}{\partial v} X \right) - \left( \frac{d}{dt} \frac{\partial L}{\partial v} \right) X
$$
$$
\frac{\partial L}{\partial t} T + L \dot T = \frac{d}{dt}\left( LT \right) - \frac{\partial L}{\partial x} \dot q T - \frac{d}{dt}\left( \frac{\partial L}{\partial v} \dot q T \right) + \left( \frac{d}{dt}\frac{\partial L}{\partial v} \right) \dot q T + \frac{\partial L}{\partial v}\dot q \dot T
$$
we arrive at
$$
\delta S = \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial v} \right)(X - \dot q T) + \frac{d}{dt} \left( \frac{\partial L}{\partial v}(X - \dot q T) + LT \right) \right]\,dt
$$
The first term vanishes if we asume the Euler-Lagrange equations, the second term yields our conservation law once we move $K$ to this side of the equation. This concludes the proof. $\square$
Note the change of region of intergration in the first step. The new time coordinate was by no means a dummy variable - the transformation is 'active', a one-parameter group of diffeomorphisms.
