The easiest way of solving such problems is to use complex phasors. The total complex impedance is
$$Z_T=R_1+R_2||j\omega L=R_1+\frac{j\omega R_2L}{R_2+j\omega L}=877.18\cdot e^{30.67^{\circ}}\Omega$$
with \$\omega=2\pi\cdot 10,000 Hz\$, \$R_1=470\Omega\$, \$R_2=988\Omega\$, and \$L=10mH\$ (as shown in your circuit diagram).
This gives for the total current
$$I_s=\frac{V_0}{Z_T}=9.12\cdot e^{-30.67^{\circ}} mA$$
with \$V_0=8V\$.
The voltage across \$R_1\$ is
$$V_1=I_sR_1=4.28\cdot e^{-30.67^{\circ}} V$$
The voltage across \$R_2\$ and \$L\$ is
$$V_2=I_s(R_2||j\omega L)=I_s\frac{j\omega R_2L}{R_2+j\omega L}=4.83\cdot e^{26.88^{\circ}} V\quad(=V_0-V_1)$$
The current through \$R_2\$ is
$$I_2=\frac{V_2}{R_2}=4.89\cdot e^{26.88^{\circ}} mA$$
The current through the inductor is
$$I_L=\frac{V_2}{j\omega L}=7.70\cdot e^{-63.12^{\circ}} mA\quad(=I_s-I_2)$$
I realize that there are quite a few differences between these results and your calculations and measurements. I just used the values you gave in the circuit diagram, and I did my best to avoid any errors.
The inductor starts discharging the voltage across the capacitor. At some point later that voltage is zero but the current is not zero. The energy in the capacitor due to its initial voltage is totally transferred to the inductor in the form of current.
Now we have the reverse situation. That inductor energy starts pumping back into the capacitor and, after the same period of time, the energy is all taken back to the capacitor in the form of a reversed voltage. The cycle repeats.
Go look up the energy equations for a capacitor and inductor and try and visualize this. Also look up the base equations that relate current and voltage to inductors and capacitors respectively.
You can also think of it like a flywheel tethered with an elastic rope. You can turn the flywheel so as to build up tension in the rope. Then let go. The flywheel starts turning the opposite way and at some time later the elastic rope has no tension in it. Because the flywheel has mass it continues spinning but is slowed down by the rope re tensioning. Eventually, the flywheel stops but, because there is energy in the rope the flywheel starts building up speed in the opposite way.
The equations are pretty much the same for this mechanical analogy and the LC circuit.
Best Answer
Don't think of the back emf as being caused by the source voltage. Think of the back emf as being caused by the changing current.
The source produces a current. That (changing) current causes the inductor to produce a back emf. The back emf limits (but doesn't eliminate) the current produced by the source. If it eliminated the current, then there would be no back emf. The two things don't oppose each other so much as they balance each other.
By KVL, the back-emf of the inductor must be be equal and opposite to the voltage produced by the source, or else energy conservation would be broken.
There is, in the sense that if you follow the path of the circuit and add up all the voltages you pass (the source and the inductor, in a simple case), then the sum of all those voltages will be zero once you've followed the complete circuit.
But of course that's just KVL, and it applies to any circuit elements (resistors, capacitors, or whatever) and not just to inductors.
I think what's confusing (at least to me) is the use of the word "opposite". In a normal drawing of this circuit, we draw the source with its positive terminal at the top of the page, and then the inductor will also produce a back emf that's positive at the top of the page. But if you consider the voltage in the direction of traversing the circuit (for example, always going in a clockwise direction around the circuit), then the two potential differences (source emf and inductor back-emf) will be opposite.
edit RE the video
I think the video is confusing because he never defines the sign convention he is using to define his voltages across the different elements.
In order for the description to be correct, he must have defined opposite conventions for the two types of elements, like this:
simulate this circuit – Schematic created using CircuitLab
Keep in mind Kirchoff's voltage law. With these conventions, \$V_r\$ must be equal to the V1 source voltage (regardless of what type of resistor is used or even if you replaced the resistor with a different element like a capacitor or inductor). And \$V_l\$ must be the opposite of the V2 source voltage (again, it would still be opposite even if you replaced the inductor with a different element like a resistor or capacitor.
Again, this is simply because of KVL and the choice of sign convention for the voltage across the two elements. But in the video he never shows the chosen sign convention, so this makes it impossible to understand what he's trying to teach.
Either that or the guy in the video has no understanding of circuit theory and believes that a circuit with an inductor in it can violate KVL, which is simply not correct.