What is the relevance of seeing $\text{ad}$ as the differential of $\text{Ad}$ $\DeclareMathOperator{\ad}{ad}$$\DeclareMathOperator{\Ad}{Ad}$
There are essentially two ways of defining the $\ad$ operator associated to a Lie Group.
The first one is purely theoretical Lie-algebra theoretical, and it's straightfoward, is just defining $\ad(X)(Y)$ as $[X,Y]$, which is readily seen to be a Lie algebra homomorphism into $\mathfrak{gl}(\mathfrak{g})$.
This operator is of course very important, as is allows us to extract some information on $\mathfrak{g}$ from is image in $\mathfrak{gl}(\mathfrak{g})$, which allows us to use for instance Jordan decompositon on matrices and such.
Of course when we have a Lie group, we have a Lie algebra structure on its tangent space at identity (or on the left invariant tangent fields), and we need to prove that we have a Lie Bracket which amounts usually to define it as the derivative at $0$ of $e^{tX}e^{tY}e^{-tX}e^{-tY}$.
However, usually we define the $\ad$ representation in the following manner: we have a morphims $\Ad:G\to GL(\mathfrak{g})$ defined to be $g\mapsto d\text{Conj}(g)(1)$ where $\text{Conj}(g):G\to G$ is the conjugation. Then $\ad$ is simply the differential of $\Ad$
Is there any advantage of definig $\ad$ like that instead of simply definig it as $\ad(X)=[X,\bullet]$.
We can prove of course that it is an equivalent definition, but what is to be gained from defining $\ad$ in this rather convoluted fashion ?
All the theorems that I know about Lie algebra (the ones in Humphreys book for instance), use the fact that $\ad$ is the Lie bracket, and not the fact that is the differential of $\Ad$.
Maybe this is important, to descend Lie algebra representations into actual representations of $G$?
 A: (Expanding on my answer in the comments)
Are you sure about your proposed alternate definition of the Lie bracket? We should have something like $e^{tX}e^{tY}e^{-tX}e^{-tY}\approx (1+tX)(1+tY)(1-tX)(1-tY)=1+t^2[X,Y]+o(t^3).$ So you can't get the Lie bracket by taking the first derivative and setting $t=0$.
I think instead you can define it as $\frac{1}{2} \frac{d^2}{ds\,dt}[e^{sX},e^{tY}]$ or $\frac{d}{dt}[e^{\sqrt{t}X},e^{\sqrt{t}Y}]$. Either way that expression is more complicated and less natural, it's harder to motivate.
And defining the Lie bracket in terms of the exponential map requires quite a bit more machinery, a discussion of uniqueness of solutions to differential equations, whereas the definition in terms of the adjoint map requires only the differential.

Is there any advantage of definig ad like that instead of simply definig it as ad()=[,∙]?

Seems like a different question. If you only care about the Lie algebra, then you may define the adjoint map solely in terms of its axiomatically given Lie bracket.
But if you want to relate the Lie bracket on a Lie algebra to the group operation on the Lie group of which it is the tangent space, then you need an equation like $[X,Y]=\frac{1}{2} \frac{d^2}{ds\,dt}[e^{sX},e^{tY}]$ or $[X,Y]=d(\text{Ad}(X))(Y).$
Ultimately the first thing you will want to prove about this relation is the functoriality of the Lie algebra functor, $\text{Lie}(G)$ which associates to a Lie group its tangent space endowed with a Lie bracket. The definition of the Lie bracket in terms of the adjoint map makes this check more straightforward than the definition in terms of exponentials.
And the adjoint map has the benefit of giving you a representation of both the group and the algebra.
So to sum up, the definition in terms of the adjoint map is actually simpler, not more convoluted, than the one you propose.
