I) The geometric argument is clear: Consider a Lagrangian density ${\cal L}=d_{\mu}F^{\mu}$ that is a total divergence. The corresponding action
$$S[\phi] ~=~ \int_M \! d^nx~{\cal L}~=~ \int_{\partial M} \! d^{n-1}x~(\ldots)\tag{0}$$
will then be a boundary integral, due to the divergence theorem. Therefore the corresponding variational/functional derivative,
$$ \frac{\delta S}{\delta\phi^{\alpha}(x)}\tag{1}$$
which is an object living in the bulk (rather than on the boundary), can never be other than identically zero in the bulk
$$ \frac{\delta S}{\delta\phi^{\alpha}(x)}~\equiv~0,\tag{2}$$
if it exists. (Note: Even for a sufficiently smooth Lagrangian density ${\cal L}$, the existence of the functional derivative is a nontrivial issue and tied to whether or not consistent boundary conditions are assumed in the variation.)
Next, recall that (the expression for) the field equations of motion is just simply given by the functional derivative (1) of the action. Then according to eq. (2), (the expression for) the field equations of motion vanish identically.
II) Finally, extend the above argument from section I by linearity to a general Lagrangian density of the form ${\cal L}+d_{\mu}F^{\mu}$ that include an extra total divergence term. Conclude via linearity, that the latter does not contribute to the field equations of motion.
I) In general, it is true that if we plug a local Lagrangian
$$\tag{1} L\quad \longrightarrow \quad \tilde{L}~=~L+\frac{df}{dt}$$
modified with a total derivative term into the Euler-Lagrange expression
$$\tag{2} \sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial \tilde{L}}{\partial q^{(n)}}~=~\sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial L}{\partial q^{(n)}}, $$
it would lead to identically the same Euler-Lagrange expression without any restrictions on $L$ and $f$.
II) The caveat is that the Euler-Lagrange expression (2) is only$^1$ physically legitimate, if it has a physical interpretation as a variational/functional derivative of an action principle. However, existence of a variational/functional derivative is a non-trivial issue, which relies on well-posed boundary conditions for the variational problem. In plain English: Boundary conditions are needed in order to justify integration by parts. See also e.g. my related Phys.SE answers here & here.
III) A Lagrangian $L(q,\dot{q},\ldots, q^{(N)},t)$ of order $N$ leads to equation of motion of order $\leq 2N$. Typically we require the Lagrangian $L(q,\dot{q},t)$ to be of first order $N=1$. See e.g. this and this Phys.SE posts.
IV) Concretely, let us assume that we are given a first-order Lagrangian $L(q,\dot{q},t)$. If one redefines the Lagrangian with a total derivative
$$\tag{3} \tilde{L}(q, \dot{q}, \ddot{q}, t)~=~L(q, \dot{q}, t)+\frac{d}{dt}f(q, \dot{q}, t), $$
where $f(q, \dot{q}, t)$ depends on velocity $\dot{q}$, then the new Lagrangian $\tilde{L}(q, \dot{q}, \ddot{q}, t)$ may also depend on acceleration $\ddot{q}$, i.e. be of higher order.
V) With a higher-order $\tilde{L}(q, \dot{q}, \ddot{q}, t)$, we might have to impose additional boundary conditions in order to derive Euler-Lagrange equations from the principle of a stationary action by use of repeated integrations by parts.
VI) It seems that Prof. V. Balakrishnan in the video has the issues IV and V in mind when he spoke of 'putting further conditions' on the system. Finally, OP may also find this Phys.SE post interesting.
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$^1$ Here we ignore derivations of Lagrange equations directly from Newton's laws, i.e. without the use of the principle of a stationary action, such as e.g. this Phys.SE post, because they usually don't involve redefinitions (3).
Best Answer
Let us for simplicity consider just classical point mechanics (i.e. a $0+1$ dimensional world volume) with only one variable $q(t)$. (The generalization to classical field theory on an $n+1$ dimensional world volume with several fields is straightforward.)
Let us reformulate the title(v1) as follows:
In short, it is because:
In physics, the action functional $S[q]$ should be local, i.e. of the form $S[q]=\int dt~L$, where the $L$ is a function of the form $$L~=~L(q(t), \dot{q}(t), \ddot{q}(t), \ldots, \frac{d^Nq(t)}{dt^N};t),$$ and where $N\in\mathbb{N}_{0}$ is some finite order. (In most physics applications $N=1$, but this is not important in what follows. Note that the Euler-Lagrange equations get modified with higher-order terms if $N>1$.)
Similarly, we demand that $F$ is of local form $$F~=~F(q(t), \dot{q}(t), \ddot{q}(t), \ldots, \frac{d^{N-1}q(t)}{dt^{N-1}};t),$$ We stress that $L$ and $F$ only refer to the same time instant $t$. In other words, if $t$ is now, then $L$ and $F$ does not depend on the past nor the future.
The special intermediate role played by the $q$ variable in between $L$ and $t$. Note that there can be both implicit and explicit time-dependence of $L$ and $F$.
Counterexample: Consider
$$L~=~-\frac{k}{2}q(t)^2.$$ Then we can write $L=\frac{dF}{dt}$ as a total time derivative by defining
$$F=-\frac{k}{2}\int_0^t dt'~q(t')^2. $$
($F$ is unique up to a functional K[q] that doesn't depend on $t$.) But $F$ is not on local form as it also depends on the past $t'<t$.
Finally, let us mention that one can prove (still under assumption of locality in above sense plus assuming that the configuration space is contractible, due to an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 1) that
$$ \text{The Lagrangian density is a total divergence} $$ $$\Updownarrow$$ $$\text{The Euler-Lagrange equations are identically satisfied}. $$
References: