I don't get why for $u$ substitution they sub the upper and lower bounds
into $u$ from the original function to find the new upper and lower
bounds with the function $u$.
[...]
My question is what or how does plugging your old lower and upper
bound values into $u$ give you the new values of your new function that's
expressed as $u$...
The blunt answer is "That's what the change of variables theorem says."
But here's a more conceptual and notational explanation. The notation $\int_a^b f(x)\, dx$ connotes "integrating from $x = a$ to $x = b$". To emphasize this, let's write
$$
\int_{x=a}^{x=b} f(x)\, dx.
$$
Now suppose you want to evaluate
$$
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx.
$$
If you make the substitution $u = g(x)$, then $du = g'(x)\, dx$ ("by the chain rule"). If $F$ denotes an antiderivative of $f$, the preceding becomes
$$
\int_{x=a}^{x=b} f(u)\, du = F(u) \Big|_{x=a}^{x=b}.
$$
Now, it should be notationally clear that setting $u = a$ and $u = b$ does not (in general) "give the right answer": Those are not the limits specified in the original integral.
To proceed, you have two choices:
Undo the original substitution by setting $u = g(x)$, and then plug in $x = b$ and $x = a$.
Find the "new limits of integration", $u = g(a)$ and $u = g(b)$, by plugging the "old" limits $x = a$ and $x = b$ into the substitution $u = g(x)$.
The second makes notational sense because "when $x = a$, we have $u = g(a)$" and "when $x = b$, we have $u = g(b)$". It should be procedurally clear the two methods are mathematically equivalent. Computationally, the second is usually less work (as both prior respondents note); the first amounts to writing something down, then erasing it.
In symbols, either approach gives
\begin{align*}
\int_{x=a}^{x=b} f\bigl(g(x)\bigr) g'(x)\, dx
&= \int_{x=a}^{x=b} f(u)\, du && \text{Substitute $u = g(x)$;} \\
&= F(u) \Big|_{x=a}^{x=b} && \text{Antidifferentiate;} \\
&= F(u) \Big|_{u=g(a)}^{u=g(b)} && \text{Change limits;} \\
&= F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) && \text{Plug in;} \\
&= \int_{u=g(a)}^{u=g(b)} f(u)\, du. && \text{Reinterpret the fundamental theorem.}
\end{align*}
So much for the explanation; what about Real Life? The notation
$$
\int_{x=a}^{x=b} f(x)\, dx
$$
is redundant, and in practice, out of laziness^H^H^H elegance, we fall back on $\int_a^b f(x)\, dx$ (sometimes to the confusion of calculus students). But when you're learning substitution the first time, it may help to write in the variable corresponding to the numerical limits, as in:
\begin{align*}
\int_{x=0}^{x=\pi/4} -2\cos^{2}(2x) \sin(2x)\, dx
&= \int_{x=0}^{x=\pi/4} u^{2}\, du && u = \cos(2x),\quad du = -2\sin(2x)\, dx; \\
&= \int_{u=1}^{u=0} u^{2}\, du && \text{When $x = 0$, $u = 1$; when $x = \pi/4$, $u = 0$.}
\end{align*}
Given a set $S$ in a measure space and a measure $dx$, you can consider the integral
$$ \int_S f dx$$
of an integrable function $f$. For instance, we might look at
$$ \int_{[0,1]} 1 dx = 1.$$
One might pronounce this as an integral of the constant function $1$ over the interval from $0$ to $1$.
There is no way to associate a sign to the specification of the set. The set has no orientation, to borrow a term from integration on manifolds.
We recognize this as being the same as
$$ \int_0^1 1 dx = 1.$$
But this latter notation is signed, as evidenced by the natural pronunciation as the integral of the constant function $1$ from $0$ to $1$. With notation, it also makes sense to talk about
$$ \int_1^0 1 dx = -1.$$
In this sense, this latter integral is signed.
More generally, there are signed integrals over any differentiable manifold. There is unsigned differentiation over any measure space.
Best Answer
$\int f(x)\,\mathrm{d}x$ is an antiderivative. It represents any function whose derivative with respect to $x$ is $f(x)$.
$\int_0^af(x)\,\mathrm{d}x$ is a definite integral, and for any of the antiderivatives $g(x)=\int f(x)\,\mathrm{d}x$ (which incorporate a constant of integration), $$ \int_0^af(x)\,\mathrm{d}x=g(a)-g(0) $$ For example, $$ \int x^3\,\mathrm{d}x=\frac14x^4+C $$ for some constant $C$, and $$ \begin{align} \int_0^ax^3\,\mathrm{d}x &=\left(\frac14a^4+C\right)-\left(\frac140^4+C\right)\\ &=\frac14a^4 \end{align} $$ no matter which $C$ is chosen.
In the case above, $\int_0^xf(x)\,\mathrm{d}x$, there is confusion because the same variable is used inside the integration as in the bounds. The bound variable $x$ inside the integral is not the same as the free variable $x$ in the limit. To reduce the confusion, your integral can also be written as $\int_0^xf(t)\,\mathrm{d}t$ by renaming the bound variable. In any case, this is a definite integral.