You wrote it yourself: it is because the domain of $f^{-1}$ is the range of $f$. Don't forget defining a function consists in defining, not only a computational formula, but its domain of validity, and its target.
So, although $\dfrac{x^2-5}2$ is a perfectly valid function defining formula for all $x\in\mathbf R$, it does not define, per se, the inverse of $f$. If you consider it defines a function with domain $\mathbf R$, this function is a continuation of $f^{-1}$, which is defined on $\mathbf R^+$, to the whole of $\mathbf R$.
By definition of increasing and decreasing,
$$ f(x) \text{ is increasing } \longleftrightarrow\frac{df}{dx}>0\\
f(x) \text{ is decreasing } \longleftrightarrow\frac{df}{dx}<0
$$
Therefore, $\frac{df}{dx}>0$ and $\frac{d}{dx}(fg)<0$. By the product rule, $f\frac{dg}{dx}+g\frac{df}{dx}<0\to\frac{dg}{dx}<-\frac gf\frac{df}{dx}$. By what we have been given, $f(x)$ and $\frac{df}{dx}$ are both positive, so as long as $g(x)$ is positive for every $x$ in your interval, $\frac{dg}{df} < 0$, which means $g(x)$ is decreasing. If $g(x)<0$ for any $x\in[0,1]$, then your conclusion is false.
Easiest counter example, let $f(x)=x$ and $g(x)=x-2$. $f(x)g(x)=x^2-2x$ and $\frac{d(fg)}{dx}=2x-2$, which is negative for all $x<1$, so it's decreasing in the interval.
Added to the Question:
For $g(x)$ to be a function such that $g(x)>0$ for $x\in(0,a)$ and $g(x)<0$ for $x\in(a,1)$, $g(x)$ must necessarily go from a positive number to a negative number, which means that it has to either be decreasing for some interval containing $a$ or it must have a discontinuity.
A discontinuous counterexample was given in the comments, namely
$$g(x)=\begin{cases}x, &x<a\\x - 2, &x\geq a\end{cases}$$
This function is always increasing, so $g(x)$ must also be specified to be continuous or the conclusion that $g(x)$ is decreasing is false.
For a continuous function, $g(x)$ must be decreasing around $x=a$, but it doesn't have to be decreasing anywhere else. For example, $g(x)=(x-b)(x-a)$, where $b=1 + \frac a2$ which decreases for $x<\frac {a + b}2$ and increases for $x>\frac {a + b}2$, and $f(x)=x^3$. $\frac{d(fg)}{dx}=5x^4-4(a+b)x^3-3abx^2=5x^4-4x^3-6ax^3-3(ab)x^2$. So now, we need to find all the values of $a$ for which the derivative is less than zero.
$$\begin{align}
5x^4-4x^3-6ax^3-3abx^2 &< 0 &(\text{substitution})\\
5x^4-4x^3-6ax^3-3abx^2 &< 5x^4-4x^3-6ax^2 &(a, b, x^2 > 0)\\
x^3(5x-4-6a) &< 0 \\
5x-4 &< 6a &((x^3>0)(\forall x \in(0,1))\\
a &> \frac16 &(\max x = 1)
\end{align}$$
In other words, as long as $a>\frac16$, $fg$ will be decreasing on $(0,1)$, so we'll pick an $a$ greater than $\frac16$. We also need to find out for what values of $a$ is $\frac{a + b}2<1$, which is simply
$$\begin{align}
a + \frac a2 + 1 &< 2\\
\frac32a &< 1\\
a &< \frac23
\end{align}$$
So $a$ also has to be less than $\frac23$.
Since $f(x)$ is positive and increasing for $x\in(0,1)$, $fg$ is decreasing for $x\in(0,1)$, and $g(x)$ is a smooth function that is positive for $x<a$ and negative for $x>a$, but is increasing for $x\in\left(\frac{a + b}2,1\right)$ as long as $a\in\left(\frac16,\frac23\right)$, we have a known counterexample to the statement that $g(x)$ must be decreasing on $x\in(0,1)$.
Best Answer
Yes, decreasing ! Indeed, let take $x\leq y$. By surjectivity, $x=f(u)$ and $y=f(v)$ for a certain $u$ and a certain $v$. Then, $f(u)\leq f(v)$ by hypothesis and so $u\geq v$ because $f$ is decreasing. By bijectivity, $u=f^{-1}(x)$ and $v=f^{-1}(y)$, therefore $f^{-1}(x)\geq f^{-1}(y)$.
Q.E.D.