For any $d \geq 2$, there are hyperplanes on which $\Delta_d$ ramifies, but for $d \geq 3$, it never ramifies **completely**. I guess there are many proofs of this fact, let me give one based on projective duality.

First note that $\Delta_d$ parametrizes polynomials of degree $d$ having a multiple root. Let me give a geometric interpretation of $\Delta_d$. Let $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ be the $d$-th Vernoese embedding of $\mathbb{P}^1$. The equation $\Delta_d = 0$ gives in the dual projective space $\mathbb{P}(S^d \mathbb{C}^2)^*$ the variety which parametrizes singular hyperplane sections of $X$. This variety is known as the **projective dual** of $X$.

For any $Z \subset \mathbb{P}^N$, which is irreducible, the reflexivity Theorem tells you that $(Z^*)^* = Z$ and for any $z \in Z_{smooth}$, the tangency locus of $z^{\perp}$ with $Z^*$ is identified **as a scheme** to $\mathbb{P}(N_{Z/\mathbb{P}^N,z}^*)$.

Going back to your situation, we have $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ is smooth, its projective dual $X^* \subset \mathbb{P}(S^d \mathbb{C}^2)^*$ is the hypersurface which equation is $\Delta_d = 0$. You want to know if there exists $x \in \mathbb{P}(S^d \mathbb{C}^2)$ such that $x^{\perp} \cap X^*$ is completely non-reduced. This is equivalent to saying that the reduced space underlying the singular locus of $x^{\perp} \cap X^*$ is equal to the reduced space underlying $(x^{\perp} \cap X^*)$.

Two cases occcur:

1) if $x \notin X$, then $x^{\perp}$ is not tangent to $X^*$ (this is because $(X^*)^*= X$). Hence the singular locus of $x^{\perp} \cap X^*$ has codimension at least one in $x^{\perp} \cap X^*$ and the equation defining $x^{\perp} \cap X^*$ can't be a $k$-perfect power.

2) if $x \in X$, then $x^{\perp}$ is tangent to $X^*$ (this is again because $(X^*)^*= X$). Since $X$ is smooth, the reflexivity Theorem ensures that the tangency locus of $x^{\perp}$ with $X^*$ is **scheme-theoretically** a hyperplane in $x^{\perp}$. As a consequence, the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane. Now, if $x^{\perp} \cap X^*$ was a $k$-perfect power, then the fact that the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane in $x^{\perp}$ would imply that $k=2$ and $\deg \Delta_d =2$. This is only possible if $d=2$.

As far as non-complete ramification is concerned, the above argument shows that for any $x \in X$, the equation of $x^{\perp} \cap \{\Delta_d = 0\}$ can be written as $L^2Q$ where $L$ is a linear factor and $Q$ a polynomial of degree $2d-4$ which is not a power.