@FakeMod This is not the most consistent example, but it has the idea: If $\mathbf{A} = x^1 \mathbf{e}_1$ is your vector, and you double the basis vector $\mathbf{e}_1$ to $\mathbf{e}_1' = 2 \mathbf{e}_1$, then $x^1$ must transform into $x'^1 = \frac{1}{2} x^1$ so that $\mathbf{A}' = x'^1 \mathbf{e}_1' = \frac{1}{2} x^1 (2 \mathbf{e}_1) = x^1 \mathbf{e}_1 = \mathbf{A}$ holds.

Here the components $x^1$ transform 'oppositely' to how $\mathbf{e}_1$ transforms, so you say the components transform 'contravariantly' to how the basis transforms. If $\mathbf{A} = x^1 \frac{\partial }{\partial x^1}$ is your vector, and you do a basis transformation double the basis vector,

in this case amounts to a coordinate transformation from $x^1$ to $x'^1 = \frac{1}{2} x^1$ so that $\frac{\partial }{\partial x'^1} = \frac{\partial }{\partial \frac{1}{2}x^1} = 2 \frac{\partial }{\partial x^1}$, then the coordinates must transform in the same way that the basis transforms,

so that $\mathbf{A}' = x'^1 \frac{\partial }{\partial x'^1} = \tfrac{1}{2} x^1 \frac{\partial }{\partial \tfrac{1}{2}x^1} = \frac{\tfrac{1}{2}}{\tfrac{1}{2}} x^1 \frac{\partial }{\partial x^1} = x^1 \frac{\partial }{\partial x^1} = \mathbf{A}$. In this case the components transform covariantly with the basis. The video gives an example using 2D projections which is better.