First and foremost I am sorry for this; in order to make the questions clear I must first upload the lecture notes from my institution $^\zeta$ for the derivation of Hydrostatic Equilibrium:
$^\zeta$ Lecture notes courtesy of Imperial College London, Astrophysics dept, 2017-2018 edition.
Firstly, in equation $(3.1)$ why is there no negative sign for $F_g$? Last time I checked, Newton's law of gravitation told us that
$$F=-\frac{GMm}{r^2}$$
Nextly, I have a conceptual problem with equation $(3.2)$; when they use the term 'pressure' I am going to assume that they are referring to the radiation pressure that is released due to nuclear fusion reactions taking place in the sun. In that formula it must be the case that $P(r) \gt P(r+\delta r)$ if this is the case then $(3.2)$ should read $$F_p=-\left(\frac{dP}{dr}\right)\delta r\,\delta A $$
Correct me if I'm wrong but Hydrogen nuclei are fused together to generate the more stable Helium and hence lots of radiation pressure acting outwards from the centre of the star. If this is correct then $P(r+\delta r)=0$, since the outside face of that cylindrical volume element in Figure $3.1$ will not 'feel' any of this radiation pressure because this pressure only acts on the inner face. What am I not understanding here?
Moving on to equation $(3.3)$, let's suppose the sign of the gravitational force in $(3.1)$ really is neglected. Does this mean that the negative sign in $(3.3)$ is due to $P(r+\delta r) \lt P(r)$ and hence $$\frac{GM(r)\rho(r)}{r^2}=-\left(\frac{dP}{dr}\right)\,?$$ I know this is the same equation as in $(3.3)$ but I have written it with the negative sign next to the pressure differential as I would like to know where this minus sign originates from. Does anyone know?
My final concern is the sign in $(3.4)$ as last time I checked $$g=-\frac{GM}{r^2}$$ and not $$g=\frac{GM}{r^2}$$
Hence, equation $(3.4)$ should read $$\left(\frac{dP}{dr}\right)=\rho(r)\,g\,?$$
I forgot to mention that I have done some research and this is the closest derivation I could find, which, needless to say is no less helpful than the notes I have.
Answer
Equation 3.1 probably should have a minus sign in front of it. The gravitational force acts in the opposite direction to which $r$ increases. However, the text clearly states that it acts towards the centre and so long as we bear that in mind there is not necessarily any problem with what is written - it is merely the magnitude of the gravitational force.
The pressure is the pressure. There is no need to distinguish what the source of the pressure is (thermal, degeneracy, radiation, magnetic fields); hydrostatic equilibrium applies in all (equilibrium) situations. In a star like the Sun the radiation pressure is negligible, but it becomes much more important in more massive stars.
In all cases $P(r) > P(r + \delta r$), so $dP/dr$ is negative and indeed the force due to the pressure gradients should be written as $$ F_P = -\left(\frac{dP}{dr}\right) \delta r \delta A,$$ if we define positive forces as acting towards increasing $r$. I think this is a problem with your notes, because the direction defined for $F_P$ is ambiguous.
The final piece of the puzzle is to sum $F_g$ and $F_P$ (as vectors) and equate to zero (for equilibrium). Since they both should have minus signs in front of them (IMO), you end up with equation 3.3. Your notes have fudged the issue and said $F_g = F_P$, which is incorrect as they are defined, and then inserted an unexplained additional minus sign to correct the mistake and you end up with the same thing.
Finally, $g$ is usually described as a positive quantity unless written as a vector $\vec{g}$, though ideally the forces might also have been written as vectors. In other words $g$ here is a scalar quantity, the magnitude of the gravitational acceleration. Thus we say the gravitational acceleration at the surface of the Earth for example, is 9.8 m/s$^2$ acting downwards, where $g = GM/r^2$.
The whole exercise illustrates the usefulness of vectors in addressing problems.
$$\vec{F_g} = -\frac{GM(r)\delta m}{r^2}\ \hat{r}$$ $$\vec{F_P} = -\left(\frac{dP}{dr}\right) \delta r \delta A\ \hat{r}$$ $$\vec{F_P} + \vec{F_g} = 0$$ leads to the scalar equation of hydrostatic equilibrium $$ \frac{dP}{dr}=- \rho g,$$ where $\vec{g}= -g\hat{r}$.
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