As per @Sudix's answer above, the intuition behind the use of the Law of Excluded Middle in the proof of :
> $(φ → ∃x ψ) ⊢ ∃x (φ → ψ)$
is to apply a "case analysis".
_(i)_ Assume that $φ$ does not hold, i.e. assume $¬φ$.
This means (by the truth-table for the _conditional_ ) that $φ → ψ$ is TRUE, and thus also $∃x (φ → ψ)$ is TRUE.
_(ii)_ Now assume that $φ$ holds, i.e. assume $φ$.
We know that the premise $(φ → ∃x ψ)$ holds, and this means (again by the truth-table for the _conditional_ ) that also $∃x ψ$ is TRUE, i.e. that $ψ$ is TRUE for some $x$.
Thus, $φ → ψ$ is TRUE for some $x$, i.e. $∃x (φ → ψ)$ is TRUE.