Is a substitution always required to change the limits of an integral?
The fundamental reason to change the limits of integration is that the variable of integration has changed.
Substitution is an obvious case in which this is likely to occur. For example, substitute $u = x - 2$ in $\int (x - 2) dx$: $$ \int_0^2 (x - 2) dx = \int_{-2}^0 u\; du. $$ The intuition I follow on this is that the start of the integral occurs "when $x=0$" and ends "when $x=2$". But "when" $x=0$, it must also be true that $u=-2$, and "when" $x=2$, it must also be true that $u=0$. So in terms of $u$, the integral needs to start "when $u=-2$" and end "when $u=0$".
A more rigorous treatment would take $x - 2$ as a function over the domain $[0,2]$ and transform it; but transforming the function also transforms its domain, so $x - 2$ over the domain $[0,2]$ transforms to $u$ over the domain $[-2,0]$.
Anything that changes the integration variable of a definite integral also has to be reflected in the limits of integration, because just as with any substitution, you're integrating a (possibly) different function over a (possibly) different domain. That is, if the integrand changes from $f(t)dt$ to $h(v)dv$ (even if $h(v)$ is a constant function, as it is in the question), the integral over $v$ needs to start and end at $v$-values that are correctly matched to the $t$-values at which the integral over $t$ started and ended.
Note that in any change of variables, regardless of whether we achieve it by first writing down an explicit substitution formula (such as $u = x - 2$), has to account for the derivative of the new variable of integration with respect to the old one. For a substitution from $t$ to $u$ via the equation $u = h(t)$, the derivative of the new w.r.t. the old is $\dfrac{du}{dt} = h'(t)$ and it is accounted for in the rule $$ \int g(h(t))\, h'(t)\, dt = g(u)\, du. $$ As far as I know, a change of variables must not break this rule, so there must somehow be a substitution $u = h(t)$ that can explain it.
In the integral in the question, $f(t) = m \dfrac{d^2 x}{dt^2}$. If $\dfrac{dx}{dt} = v = h(t)$ and if $g$ is the constant function with value $m$, then $\dfrac{d^2 x}{dt^2} = \dfrac{dv}{dt} = h'(t)$ and $g(h(t)) = m$, so $$ \int m \frac{dv}{dt} \, dt = \int g(h(t))\, h'(t)\, dt = \int g(v)\, dv = \int m \, dv. $$ The definite integral follows the same rule but also has to make the corresponding change to the interval of integration: $$ \int_{t_0}^{t_1} g(h(t))\, h'(t)\, dt = \int_{h(t_0)}^{h(t_1)} g(v)\, dv; $$ setting $v_0=h(t_0)$ and $v_1=h(t_1)$, $$ \int_{t_0}^{t_1} m \frac{dv}{dt} \, dt = \int_{h(t_0)}^{h(t_1)} m \, dv = \int_{v_0}^{v_1} m \, dv. $$
Apart from your prefactors (like mass), when you integrate $dv/dt$ you integrate the derivative of a function, hence we all know the answer is the function itself. Hence: $$\int_{t_0}^{t_1} v^{\prime}(t)\, dt = v(t_1) - v(t_0)$$