How does an Indefinite Integral = Definite Integral with Variable as Upper Limit? Source: P36 of Elementary Differential Equations, 9th Ed by Boyce, DiPrima et al.

$${\int{f(t)\text{ }dt} = \int_{t_0}^t f(s) \text{ } ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}}  \tag{$*$}$$

$\Large{\text{1.}}$ I now know: 
$ \int{f(t)\text{ }dt} \qquad$ is $\color{green}{\text{  a set of functions $\qquad$ (**)}} $
and  $  \int_{t_0}^t f(s) \text{ } ds \qquad$ is $\color{#B53389}{\text{ an element of set (**) above.}} \quad $
So how and why is $(*)$ true? How can a $\color{green}{\text{  a set of functions}}$ = $\color{#B53389}{\text{ an element of the same set}}  $  ?   

Supplementary to William Stagner's answer and Peter Tamaroff's comment
Thanks to your explanations, I now know that: 
$\int{f(t)}\text{ }dt = g(t) + C \qquad \forall \ C \in \mathbb{R}\ \tag{$\natural$}$
$\int_{t_0}^t f(s) \text{ } ds = g(t) - g(t_0) \tag{$\blacklozenge$}$
Since $g(t)$ is one function and $t_0$ is one arbitrarily chosen argument/number,
thus $-g\left(t_0\right)$ is ONE FIXED number.
In contrast, $C$ is ANY real number.
$\Large{\text{3.}}$ So $(\natural) \mathop{=}^{?} \, (\blacklozenge) \iff C \mathop{=}^{?} -g(t_0).$ But how and why is : $ C \mathop{=}^{?} -g\left(t_0\right) \; $?    
 A: On the LHS of the grey box, $\int{f(t)\,dt}$ is the antiderivative of $f$.
It is defined up to a constant: $\int{f(t)\,dt} = g(t) + C.$
On the RHS of the grey box, $\int_{t_0}^t f(s) \, ds$
is the definite integral of $f(x)$. 
By the fundamental theorem of calculus, $\int_{t_0}^t f(s) \, ds = g(t) - g(t_0).
$
Two expressions are equal when $C = - g(t_0)$. Formally, $\int{f(t)\,dt}$ is a set of functions of the form $f(x) + C$, and $\int_{t_0}^t f(s) \, ds$ is an element of this set.
A: Since Yury answered your first two questions, I'll attempt your supplementary ones.
As he stated,  $$\int f(t)\ dt = g(t) + C$$ is well defined up to a constant. What does this mean? There is a whole class of antiderivatives of $f$ which we can specify exactly, namely $g(t)$ plus some constant. Thus, the set of anti derivatives of $f$ is $$\{ g(t) + C : C\in \mathbb{R}\}.$$ So we're looking at a set of functions. The expression $g(t) +C$ is really just shorthand for the set of functions defined above. Hence, $\int_{t_0}^t f(s)\ ds = g(t) - g(t_0)$ is just an element of this set, where $C=-g(t_0)$. 
I think this answers both of your questions. The expression $g(t) + C$ is not the antiderivative of $f$, but represents the set of antiderivatives of $f$, which differ from $g$ by the addition of a constant.
Question 3
If I understand correctly, this is mainly an ambiguity between fixed and free variables. It is not true that "we don't know the value of $t_0$." We do know the value, it is in fact, $t_0$! Although $t_0$ is arbitrary, it is still a fixed variable. So in the expression
$$
g(t) - g(t_0)
$$
the variable $t$ is free to vary over $\mathbb{R}$, but $t_0$ is absolutely fixed from the beginning (although arbitrary).
Does this clear things up?
Also, strictly speaking, it would be more accurate to say that $(**) \in (*)$.
A: I edited some of rajb245's comments to unify them. ---
The question's really about interpreting what the indefinite integral symbol means. The LHS of $(*)$ is the old definition of the indefinite integral and the RHS of $(*)$ is the new definition but expressed as a definite integral. You think of the LHS of $(*)$ representing a family of curves with the same derivative, indexed by a parameter $C$ . If you let $t_0$ be arbitrary too, the RHS of $(*)$ gives a family of curves as well. These curves are indexed by $t_0$. Generally, the RHS of $(*) \subseteq$ LHS of $(*)$. 
However, if you DEFINE the indefinite integral sign to have the property $(*)$, then for arbitrary but fixed $t_0$, RHS of $(*)$ generates a subset of the functions under the LHS of $(*)$. At most, you lose some of the antiderivatives. This RHS of $(*)$ has more useful properties than the more general $+C$  form.
Another way to fathom this ----
Suppose you knew nothing of indefinite integration, and the symbol had no meaning to you. And suppose you understand definite integrals in terms of convergent sequences of Riemann sums. Now someone defines the indefinite integral to you in terms of the RHS of $(*)$. 
Then question 1.1 presupposes that the symbol on the left represents an entire set. But under the way of thinking that I'm proposing, we never established this at all.
As for your text not introducing $(*)$ as a definition, then the answer is that they were sloppy to introduce the anti-derivatives first. 
Another way of thinking of $(*)$ is to interpret the equality given as: "An antiderivative of $f(t)$ is given by the following convergent limit of a sequence: ..." This is absolutely a true statement.
