Why do pseudoforces always act in opposite direction to acceleration? Why are the pseudoforces felt by bodies inside an accelerated frame of reference opposite to the acceleration of the frame? I know that it is true by observation, but would wanna know why exactly. Is it to do with inertia, since the body inside the frame of reference is being forced to accelerate from its initial state of rest/motion? That is the best explanation that comes to mind
 A: It is due to an object's inertia; which is a measure of how "hard" it is to accelerate.
Since inertia acts to "dampen" acceleration, the resulting pseudo force acts in the opposite direct to acceleration.
A: Let's try to explain it with a rough example.
Imagine you're in a bus which suddenly accelerates.
Someone outside the bus would see "Well, obviously the bus has an acceleration, but the person inside does not, so applying NEwton's laws:
$bus:\ F=ma$ The bus will accelerate forwards.
But there is no motor force on the person, so $person:\ F=0 \Rightarrow a=0$
The person will stay still while the bus goes forwards.
Fortunately, friction force will make you accelerate progressively like the bus, reducing relative motion between you and the platform.

Now, from the passenger's point of view, the passenger sees he's goign backwards. Yet there is no force acting on the person, so how can a person be accelerating backwards without forces acting on him?
The reason is that Newton laws did not provide for non inertial frames...
Consequently, the passenger cannot use Newton's laws ordinarily. If he did, he'd find total force $F=0$ but $a$ backwards.
But we want to keep using Newton's laws, and we do that by adding a new force, a fictious one. He dows that by asking the outsider: Hey, it seems I am a non inertial frame because I see Newton's law failure. Can you lpease tell me with what acceleration I am moving respect to you?
And then, by knowing $a_{bus}$, he can correct Newton's laws.
AS the other asnwer says:
$a_{inertial}=a_{bus}+a_{nonI}$
Multiply all by the mass...
$$ma_{inertial}=ma_{bus}+ma_{nonI}$$
$$F_{nonI}=F_{inertial}-ma_{bus}$$
So, if the passenger wants to keep using Newton's laws, he must add a new correction term, a fictious force. The passenger sees the same forces as the inertial one, plus one correction term. And it is negative because of Galilean transforms.
A: Lets set up the notations:

*

*$\vec{r}$ is position vector of point P referred to an rest inertial frame O.

*$\vec{r'}$ P referred to the frame A (non-inertial).

*$\vec{\xi}$ the origin of frame A measured in frame O.

We have vector relation:
$$ \tag{1}
  \vec{r} = \vec{\xi} + \vec{r}'.
$$
The velocity of point P from differential Eq.(1)
\begin{align}
  \vec{v} =& \vec{V} + \vec{v}'.\\
  \vec{v} =& \frac{d\vec{r}}{dt}; \,\,\, \vec{v}' = \frac{d\vec{r}'}{dt};\,\,\, \vec{V} = \frac{d\vec{\xi}}{dt};
\end{align}
The accelerations:
\begin{align} \tag{2}
  \vec{a} =& \vec{A} + \vec{a}'.\\
  \vec{a} =& \frac{d^2\vec{r}}{dt^2}; \,\,\, \vec{a}' = \frac{d^2\vec{r}'}{dt^2};\,\,\, \vec{A} = \frac{d^2\vec{\xi}}{dt^2};
\end{align}
In inertial frame O, the $\vec{F} = m\vec{a}$ applied. Therefore, observes in frame A:
$$
 m\vec{a}' = m\vec{a} - m \vec{A}
$$
In terms of force
$$
 \vec{F}' = \vec{F} - m \vec{A}.
$$
The last term is what we called  fictitious force.
A: It is an abuse of language in my opinion. For example, if our car makes a curve to the right, our body tends to move to the left. So we intuitively postulate a force to the left to explain that movement.
But we don't feel any force really. We simply observe that we are moving to the left. We only feel a force when touching something (the door for example). And in this case the force acting on us is to the right.
Pseudoforce is a good name because it is not something that we really feel, it is only a consequence of trying to apply the equation $F = ma$ in a situation (non inertial frame) where it is not valid.
