The basic physics in laymen's terms
Okay, so the basic idea is: an object in motion tends to stay moving at the speed that it's moving.
When we apply this to rotational dynamics we have an interesting effect: an objects speed goes linearly with the radius it is from the center it rotates around. So if something is rotating with a period T, it must go a distance $2 \pi R$ in time $T$, so its actual speed at any point is $v 2 \pi R / T$. Very often we call this parameter $2 \pi / T$ by the name "angular velocity" with symbol $\omega$ (Greek lowercase "omega"), in a small time t you sweep out an angle $\omega t$ measured in radians.
So to keep $v = R \omega$ the same, when something comes closer to the axis of rotation (lower R) it increases $\omega$ in inverse proportion, which looks like "spinning faster". This is why, for example, figure skaters pull their arms in to spin faster or let them out to spin slower.
Why that leads to twisting
Now if the gymnast leaves both arms up, he will execute just the somersault with no twist: the axis of rotation is horizontal and runs parallel to the shoulders. Lowering an arm means that the two sides of that gymnast want to rotate at different angular-speeds around this axis. And it's this behavior which manifests as a twist about the central axis of the gymnast.
Is the mathematics formidable? Actually, it's not too hard. We have to track a couple different velocities here, but to keep it simple, just imagine that we're tracking the velocities of: the center of mass in your gut, the throat right above it, and the shoulders beside that. Three points are actually enough to specify any rigid object's position and orientation; the throat we just introduce for convenience as a point halfway between the shoulders and therefore having the average of their velocities.
We first subtract off the center-of-mass motion as boring: the center of mass just follows a nice parabola, good for it, we don't care. So now we can view this person as flipping and spinning in free space far away from gravity, since those are two separate concerns. We will still think about a virtual "ground" underneath him, though, and being "horizontal" or "vertical", "up" or "down" in reference to it.
Okay so, backwards somersault, early twist. Our jumper starts out vertical, then slowly rotates onto their back, so they are horizontal and "face-up" when they suddenly pull the left arm in towards the center of the body. Right before, the shoulders and throat are moving downward with some velocity $-s$. The left side picks up an additional downward velocity $-\Delta s$. If we write (left-shoulder velocity, throat velocity, right-shoulder velocity) we would write that what's happened has been the change $(-s, -s, -s) \mapsto (-s - \Delta s, -s - \frac{\Delta s}{2}, -s)$, as the shoulder in motion wants to stay in motion and the other shoulder has picked up this added speed.
So some rotation about the original axis has "disappeared", because that's what the throat-velocity tells us. Subtract the new throat-velocity (flip-rotation) off, and you see that the remaining velocities are: $(-\frac{\Delta s}{2}, 0, +\frac{\Delta s}{2})$. So that produces a twist of the left shoulder backwards and the right shoulder forwards, which is what you'd experience, as a gymnast, as a counterclockwise rotation.
If you rotate with the somersault, the twist always looks the same
In fact, it doesn't matter which way you're facing, as far as I can tell: "adduction" (lowering) of the left limb gives counterclockwise rotation (as seen by you) for a back-somersault both at the beginning and end of your trajectory, whether you are trying to twist early or late.
However, maybe you are judging twists from the ground, particularly from a frame that is looking at the jump "in the path" that the jump is happening. Then, when you see someone who is face-up vs. face-down executing what they thing is a counterclockwise rotation, it will look counterclockwise when we see them head-on (we look down at the ground when we label our rotations as counterclockwise or clockwise, rather than looking at the sky, so it's a bird's eye view) or clockwise when we see them foot-on. So if this gymnast is coming toward you, a left-arm lowering on the first half of their jump, when their head is facing you, looks counterclockwise while on the second half of their jump it looks clockwise, and vice versa.
Fictitious forces, angular velocities
The mathematics which applies here, allowing us to describe the world from the perspective of the person who is somersaulting, is the mathematics of rotating reference frames. When you make your coordinates rotate in some way, like with the initial "backflip" of the gymnast, then you have to add two "fictitious forces". We'll say that the gymnast's center-of-mass is moving horizontally in the $\hat x$ direction, and vertically in the $\hat y$ direction. We'll also describe the direction to the gymnast's left, which as they start out is $\hat x \times \hat y$ in terms of the cross product, as the $\hat z$ direction: in this direction sits some audience watching the gymnast move from their left to right.
They start off rotating head-over-heels with some angular frequency $\omega$; in physics we treat this as a vector whose magnitude is $\omega$ but whose direction is in the axis of rotation. That specifies only two opposite directions, so the usual convention is the "right hand rule", you curl your right hand in the direction that they're going and your thumb points in the "correct" direction; if you are looking at the rotation vector $\vec \omega$ "tip-on" then the rotation looks counter-clockwise. So that means that our rotation vector for a backflip in our rotating reference frame, in terms of the ground-coordinates, is:$$\vec \omega = - \omega \hat z$$.
With the two fictitious forces, we can just do physics in the co-rotating frame. Suppose we have a rigid 3-body system with a mass M at position $(x, y, z) = (0, -d, 0)$ with two massless "shoulders" at $(x, y, z) = (0, 0, \pm s)$ supporting two "arms" of mass m at height h: $(0, h, \pm s)$. We choose $d$ so that the center of mass is at 0; this means $d = 2 m h / M$. This means that there's a force-balance of centrifugal forces $-m \vec \omega \times (\vec \omega \times \vec r)$, as well as a torque-balance: in our co-rotating reference frame, nothing is yet changing.
Two delta-function forces, some irrelevant centrifugal/center-of-mass cartwheeling.
In a short time $t$ we pull $h$ entirely into its shoulder at a constant speed $h/t$. This requires two short intense forces directed from the left arm to the shoulder, which undo each other eventually but will pull the system up by a distance $d/2$ and rotate the system a little about its center-of-mass in the "cartwheel-right" ($\hat x$) direction. We'll ignore this because $M$ is large compared to $m$.
There is now a centrifugal torque: we used to have a centrifugal torque balance and we still almost (for small $d$) have a centrifugal force balance which becomes exact when we accept the slight rotation above: but now the torque due to the centrifugal force on the right $m$ is unbalanced. This leads to a more persistent "cartwheel-left" (-\hat x) rotation, at least at first: at it goes more in this direction it decreases its own effort arm and increases the effort arm of $M$ and moves the left $m$ to counterbalance. We can imagine that it basically oscillates about an equilibrium, which we'd call precession about an axis. This is also not really the twisting.
These "cartwheel" effects are definitely somewhat noticeable; see e.g. this instructional video where you can see that, while the young man is in midair and upside-down, he is not pointed straight along the $\hat y$ axis but along some combination of $\hat y$ and $\hat z$.
Twist comes from the Coriolis force.
The remaining effect is the Coriolis force, $-2m \vec \omega \times {d\vec r \over dt}.$ This does not have a long-lasting torque because after a short time $t$ we know that $\frac{d\vec r}{dt} = 0$ for the left-mass. But while it is moving, the Coriolis force provides a constant force$$-2 m (-\omega \hat z)\times(-\frac{h}{t}~\hat y) = +2m \omega ~\frac{h}{t}~\hat x$$ to either shoulder if its arm drops/adducts: a "forward" motion stemming from both the sign of $h$ as positive and the choice of direction for $\vec\omega$ along the $-\hat z$ axis.
If you start off facing $-\hat x$ then the $-\hat z$ rotation is a "backwards somersault" and "your left" shoulder is at $s \hat z$. This motion of your left shoulder appears to be "towards your back", as I mentioned above. Let's do this explicitly.
The torque of the force is $$\vec \tau = \vec r \times \vec F = \int_0^1d\alpha~\left[\begin{array}{c}0\\ \alpha h \\ s\end{array}\right] \times \left[\begin{array}{c}2m \omega h/t\\ 0\\ 0\end{array}\right] = \left[\begin{array}{c}0\\ 2 ~m~ \omega~ h ~s / t\\-m ~\omega ~h^2/t\end{array}\right]$$which acting over a time $t$ produces a net angular momentum in the co-rotating frame of:$$\vec L = t~\vec \tau = \left[\begin{array}{c}0\\ 2 ~m~ \omega~ h ~s \\-m ~\omega ~h^2\end{array}\right]$$The z-component of this expression suggests indeed that you start rotating forward faster (because you pulled in a limb making yourself spin a little faster!) and also that you start "twisting" about the $\hat y$ axis. Your moment of inertia about this axis is just $2 m s^2$ so you should describe a rotation rate $+~\hat y~\omega~h/s$, which is counterclockwise when seen from the top down, the way that you would describe your rotation.
Again, from the ground, the issue is basically just that if you set a top on a glass spinning counterclockwise when viewed "normally" from the top down, then if you look at it from under the glass table you'll see it move clockwise. Similarly, whether you see the gymnast's rotation as clockwise or counterclockwise from the ground depends on whether their head is pointed at you (counterclockwise) or their feet are pointed at you (clockwise).
