I do not see any asymmetry in your graph. What I do see is that as the amplitude of oscillations decreases then so does (1) the period and (2) the average tension in the string ("force") during one oscillation.
(1) Although for small amplitudes $\theta_0$ (in radians) the period $T$ is approximately constant (independent of $\theta_0$), for larger amplitudes the period is longer :
$T=2\pi\sqrt{\frac{L}{g}}[1+\frac{1}{16}\theta_0^2+\frac{11}{3072}\theta_0^4+...]$....(eqn 1)
This is because the restoring force is proportional to $\sin\theta$ rather than $\theta$. When the amplitude gets to $90^{\circ}$ the period is almost $20$% above its small-angle value.
(2) The tension $F$ in the string ("force") is the sum of the component of the weight $mg$ of the bob along the string plus the centripetal force required to keep the bob moving in a circle of radius $L$ with speed $v$ :
$F=mg\cos\theta+\frac{mv^2}{L}$....(eqn 2)
Since angular displacement $\theta=\theta_0\sin(\omega t)$, force $F$ is not a strictly sinusoidal function. Here $\omega=\frac{2\pi}{T} \approx \sqrt{\frac{g}{L}}$ is the angular frequency of the pendulum. $v$ is related to displacement $\theta$ by
$v=L\frac{d\theta}{dt}$
$\frac{v^2}{L}=L\omega^2\theta_0^2\cos^2(\omega t)=g(\theta_0^2-\theta^2)$
therefore
$F=mg(\cos\theta+\theta_0^2-\theta^2)$...(eqn 3).
Note that $\omega$ is not equal to $\frac{d\theta}{dt}$. $\omega$ is related to the period of the pendulum and is constant if amplitude is constant, whereas $\theta$ is angular displacement and $\frac{d\theta}{dt}$ is angular velocity of the bob, both of which vary sinusoidally during one oscillation.
Tension is greatest in each cycle when the bob swings through the vertical position $(\theta=0)$ :
$F_{max}=mg(1+\theta_0^2)$....(eqn 4)
Tension is least in each cycle when the bob reaches the maximum displacement $(\theta=\theta_0, v=0)$ :
$F_{min}=mg\cos\theta_0$....(eqn 5)
As amplitude decreases, $F_{max}$ decreases but $F_{min}$ increases. When the pendulum stops $(\theta_0=0,v=0)$ the tension is $F=F_{max}=F_{min}=mg=1.82N$, which is the value for the blue line on your graph.
The graph above shows how the force in Newtons varies with time in seconds, starting from an amplitude of 45 degrees, with a value for damping of $\gamma=0.05$ (which is considerably higher than in your results). I have used the small-angle value for the period throughout. Note that the average force decays from about 2.1N to about 1.82N.
You have not explained the purpose of your experiment, so it is not clear how you intend to make use of your results, which show how $F$ varies with time $t$ rather than with amplitude $\theta_0$.
(i) You could investigate the damping by rearranging eqns 3 & 4 to find how $\theta_0(t)$ varies with time. The 2 eqns should give the same values for $\theta_0$; you could calculate both and use the average.
If the damping force is proportional to the speed of the bob (which is what is expected for air resistance at 'low' speeds) you should get an exponential decay $\theta_0 \propto e^{-\gamma t}$ where $\gamma$ is the decay constant. So a plot of $\ln(\theta_0)$ against $t$ should be a straight line with -ve slope.
(ii) You could investigate how the period $T$ varies with amplitude $\theta_0$, by comparing your results with the prediction in eqn 1. Again you need to obtain $\theta_0(t)$ from the values of $F_{max}$ and $F_{min}$.
It is probably best to average the period over a few oscillations. Note also that force $F$ reaches a maximum twice in each oscillation of the pendulum, so the period of the pendulum will be twice that in your graphs.
Best Answer
The difference is that the ball bearing is rolling on a circular track instead of sliding. When analyzing the motion we must take into account the moment of inertia of the ball bearing.
In a simple pendulum the bob rotates about the pivot point. This is equivalent to sliding (without friction) on the circular track. But the ball-bearing pendulum is rotating about its own centre as it also rotates about the imaginary pivot point at the centre of the circular track.
For a certain amplitude of swing, there is a fixed amount of energy. At the lowest point the kinetic energy of the ball bearing is divided between linear motion of the CM and rotation about the CM, whereas for the simple pendulum bob it all goes into linear motion of the CM. So the linear speed of the CM of the ball bearing is slower than that of the simple pendulum bob, resulting in a longer period for the ball bearing. When we come to measure the period, it is only the back-and-forth motion of the CM which we measure; we ignore the rotation of the ball bearing.
According to Oscillation of a rolling sphere in a bowl, the period of small oscillations is
$$T=2\pi \sqrt{\frac{7R}{5g}}=1.1832T_0$$ where $T_0$ is the period of a simple pendulum of the same length $R$. In the lecture $T_0=1.85s$ so we should expect $T=2.19s$. We can only assume the remaining difference with the measured value of $2.27s$ is due to other errors - eg the measurement of $R$.