What does it mean that one frequency aliases another?
I assume that since you're watching lectures on sampling theory, then you have already grasped the core ideas of basic signal operations and Fourier transforms and signal spectrum plots.

Then, consider a signal $x(t)$ whose C.T. Fourier transform is $X(\omega)$. When this signal is multiplied by the (ideal) periodic impulse train $$ \sum_{k=-\infty}^{\infty} \delta(t-kT_s) $$ where $T_s$ is the sampling period (and therefore $F_s = 1/T_s$ is the sampling frequency in Hz.) the resulting signal is called as the sampled signal $$ s(t)=x(t) \sum_{k=-\infty}^{\infty} \delta(t-k T_s) $$ 

Then it can be shown (by applying properties of CTFT) that the spectrum $S(\omega)$ of the sampled signal $s(t)$ is the periodical extension of the spectrum $X(\omega)$ of the signal $x(t)$, i.e., $S(\omega)$ is obtained by shifting $X(\omega)$ to left and right in frequency by the amount $ k \times \omega_{s} = k \times 2 \pi F_s$.  ( Here I'm using radian frequency $\omega$ which is related to Hertz frequency $f$ by $\omega=2\pi f$ ) : 
$$ S(\omega) = \frac{1}{T_s} \sum \limits_{k=-\infty}^{\infty} X(\omega -k 2\pi F_s) $$

For example consider the the shift for $k=-1$, then $X(\omega+2\pi F_s)$ is added on top of original spectrum $X(\omega)$. Now by setting $\omega=0$ in $S(\omega)$, it can be seen that $X(0+2\pi F_s)=X(2\pi Fs)$ has come to the origin in addition to $X(0)$. Setting $k=-2$ would show also that $X(4\pi F_s)$ has come to the origin. This way it can be deduced that those frequencies given by $\omega_k = k 2\pi F_s$ are all appearing at the origin in $S(\omega)$. This appearance of those frequencies $2 \pi k F_s$ at the origin (or more generally any frequency  $\omega_k = 2 \pi k F_s + w_0$ at the frequency $\omega_0$) is called as aliasing in the spectrum of $S(\omega)$