This stems from one of the "fundamental" theorems of calculus. You are asking, I think, why is it the case that whenever $F'=f$, it follows that we can compute
$$\tag{1}F(b) -F(a) = \int_a^b f(x)dx.$$
The idea is as follows. Define $G(t) = \int_a^t f(x)dx$, so this gives "the area up to a certain time $t$". Then, naturally, you are looking for $G(b)$, but (unfortunately) we do not know how to compute this.
The insight of equation $(1)$ is that the function $G$ can be differentiated, and that its derivative equals $f$. To see why this is the case, we note that
$$ G(t+h)-G(t) = \int_t^{t+h} f(x)dx = h\cdot f(\xi)$$
for some mid-point $\xi\in [t,t+h]$. As $h\to 0$, we see that $\xi\to t$, and assuming $f$ is continuous (which is the case, at least usually in first calculus courses) we get that
$$G'(t) = f(t).$$
The takeaway is that, because any two functions $F$ and $G$ with $F' = G'$ differ by a constant, we see that $F(t) - G(t)$ is constant, and this means that
$$F(b)-G(b) = F(a) - G(a)$$
and a little rearranging (plus $G(a)=0$) shows
$$F(b) - F(a) = \int_a^b f(x)dx.$$
This is an excellent question which I myself have asked quite a while ago. The answer here lies in the following proof:
Let $f(x)$ be a function continuous on the interval $[a,b]$.
Let the integral functional $\int_a^b f(x) \textrm{d}x$ denote the following expression:
$$\lim_{n\to\infty} \sum_{i=0}^{n-1} f\left(a+\dfrac{b-a}{n}i\right) \dfrac{b-a}{n}$$
Let $\Delta x=\dfrac{b-a}{n}$ and $x_i = a+i\Delta x$. This reduces the expression to
$$\lim_{n\to\infty} \sum_{i=0}^{n-1} f(x_i) \Delta x$$
This is the area under $f(x)$ from $a$ to $b$ because it's an infinite sum of rectangles of arbitrarily small length. With this in mind, let us now define $F(x)$ to be $\int_a^x f(t)\textrm{d}t$ for $x\in[a,b]$. We may now begin our proof.
Observe:
$$\dfrac{\textrm{d}F}{\textrm{d}x} = F'(x) = \lim_{h \to 0} \dfrac{F(x+h)-F(x)}{h}$$
$$\dfrac{\textrm{d}F}{\textrm{d}x} = F'(x) = \lim_{h \to 0} \dfrac{\int_a^{x+h} f(t)\textrm{d}t-\int_a^{x} f(t)\textrm{d}t}{h}$$
Based on what we know about areas and how they work, we can say that the area under $f(x)$ between $b$ and $c$ is equal to the area under $f(x)$ between $a$ to $c$ minus the area under $f(x)$ between $a$ to $b$, or $\int_b^c f(x) \textrm{d}x = \int_a^c f(x) \textrm{d}x - \int_a^b f(x) \textrm{d}x$.
This is also true because $\int_b^a f(x) \textrm{d}x=-\int_a^b f(x) \textrm{d}x$ and $\int_a^b f(x) \textrm{d}x + \int_b^c f(x) \textrm{d}x = \int_a^c f(x) \textrm{d}x$.
In any case, our original derivative equation reduces to
$$ F'(x) = \lim_{h \to 0} \dfrac{1}{h} \int_x^{x+h} f(t) \textrm{d}t $$
We may now apply the mean value theorem, which states the following:
$$\exists c \in [a,b] \textrm{ s.t. } f(c) = \dfrac{1}{b-a} \int_a^{b} f(t) \textrm{d}t$$
If we rewrite $F'(x)$ as being $\dfrac{1}{(x+h)-x} \int_x^{x+h} f(t) \textrm{d}t$, we will easily see that we can apply MVT:
$$\exists c \in [x,x+h] \textrm{ s.t. } f(c) = \dfrac{1}{(x+h)-x} \int_x^{x+h} f(t) \textrm{d}t$$
From this, we observe that $ f(c) = F'(x) $ for some $c$ between $x$ and $x+h$.
But as $h \to 0$, $x\leq c\leq x+h$ becomes the inequality $x\leq c\leq x$. And if we are to apply the squeeze theorem, we can say that the only way for this inequality to be true is if $c=x$. Therefore, $f(x) = F'(x)$. $\square$
Best Answer
In the notation of the post you linked to, you are confusing $\mathcal{M}$ and $F$. What you are calling $F(b)$, a function which measures the area under the curve from $0$ to $b$, is what the post calls $\mathcal{M}(b)$. The function $F$ is instead any antiderivative of $f$. That is, we know nothing about $F$ at all other than that it is some function whose derivative is $f$.
This doesn't mean that $F$ is the same as $\mathcal{M}$, since a function can have more than one antiderivative! Indeed, for any constant $C$, $F(x)=\mathcal{M}(x)+C$ is another antiderivative of $f$, since adding a constant does not change the derivative. However, this is the only way to get another antiderivative: if $F$ is an antiderivative of $f$, then $F'(x)-\mathcal{M}'(x)=f(x)-f(x)=0$, so the function $F(x)-\mathcal{M}(x)$ has derivative $0$ and hence is a constant.
So, we're starting with some function $F$ which is an antiderivative, but what we actually want is $\mathcal{M}$. To fix this, we have to subtract a constant from $F$, and that constant is exactly $F(0)$, since $\mathcal{M}(0)=0$.