Denote the curve class by $\beta$. Let $M$ be the normalization of a closed subvariety of $\overline{\mathcal{M}}_{0,2}(X,\beta)$ such that the restricted evaluation morphism, $$\text{ev}|_M = (\epsilon_1,\epsilon_2):M\to X\times X,$$ is surjective and generically finite. Denote by $m$ the degree of $\text{ev}|_M$. For every morphism $f:Y\to X$ as above, the claim is that $$\langle [D_f], \beta \rangle \geq 2d - 2(m!).$$

Fix a rational point, $$x_0:\text{Spec}(\mathbb{C}) \to X\setminus D_f,$$ (so I am assuming the field is $\mathbb{C}$, and I am about to start arguing geometrically, but there should be a better argument that avoids this). Denote by $M_{x_0}$ the fiber product with its remaining evaluation morphism to $S$, $$M_{x_0} = M\times_{\epsilon_2,X,x_0} \text{Spec}(\mathbb{C}) \xrightarrow{e_1} X. $$ This morphism is generically finite with a branch divisor $E_{x_0}$. Let $F$ denote $D+E_{x_0}$. Denote $X^o = X\setminus F$, and similarly $Y^o = f^{-1}(X^o)$ and $M^o_{x_0} = e_1^{-1}(X^o)$. Then the morphisms, $$f^o : Y^o \to X^o, \ \ e_1^o : M^o \to X^o, $$ are both finite and étale. Fix another rational point, $$ x_1:\text{Spec}(\mathbb{C}) \to X^o,$$ and fix points above it, $$y_1:\text{Spec}(\mathbb{C}) \to Y^o,\ \ m_1:\text{Spec}(\mathbb{C}) \to M^o.$$ There are pushforward homomorphisms, $$\pi_1(f^o): \pi_1(Y^o,y_1) \to \pi_1(X^o,x_1), \ \ \pi_1(e^o_1): \pi_1(M^o,m_1) \to \pi_1(X^o,x_1).$$ The index of the second map equals $m$, since $e_1^o$ is finite, étale of degree $m$ with normal, connected domain. The kernel of the corresponding "monodromy representation" from $\pi_1(X^o,x_1)$ to $\text{Aut}((e_1^o)^{-1}(x_1))$ is a normal subgroup $N$ of finite index $\mu$, where $\mu \leq m!$, and I write these words to avoid ending the sentence with an exclamation point.

Associated to $f^o$, there is also a monodromy representation $\rho$ of $\pi_1(X^o,x_1)$ on $(f^o)^{-1}(x_1)$, and the action is again transitive since $Y$ is normal and connected. Consider the induced action of $N$ on $(f^o)^{-1}(x_1)$. Since $\rho$ is transitive, and since $N$ has index $\mu$, there are at most $\mu$ orbits for $\rho|_N$.

For every pair $y'$, $y''$ of points of $(f^o)^{-1}(x_1)$ that are in the same orbit of $N$, there is a real continuous path $\gamma$ in $Y^o$ from $y'$ to $y''$ whose image under $f^o$ is a loop $f(\gamma)$ in $X^o$ based at $x_1$ and whose homotopy class is in $N$. In particular, the loop lifts to a loop $\gamma_M$ in $M^o$ based at $m_1$.

Denote by, $$u_{m_1}:(\mathbb{P}^1,0,1) \to (X,x_0,x_1),$$ the morphism parameterized by $m_1$. Denote by $\widetilde{C}$ the inverse image of this curve in $Y$. This decomposes into irreducible components $\widetilde{C} = \widetilde{C}_1 \cup \dots \cup \widetilde{C}_n$. By the hypotheses on $x_0$ and $x_1$, the fibers $f^{-1}(x_0)$ and $f^{-1}(x_1)$ are each reduced with $d$ elements. The decomposition of $\widetilde{C}$ into irreducible components induces partitions of $(f^o)^{-1}(x_0)$ and $(f^o)^{-1}(x_1)$ into $n$ distinct sets. Moreover, there is a natural bijection between these sets of size $n$: the partition set of an element $y$ in $(f^o)^{-1}(x_0)$ corresponds to the partition set of an element $y'$ of $(f^o)^{-1}(x_1)$ if and only if $y$ and $y'$ are in a common irreducible component of $\widetilde{C}$.

As we vary the point $m_1$ in $M^o$, the point $x_0$ and the corresponding partition of $(f^o)^{-1}(x_0)$ do not vary. Thus, if we take the loop $\gamma_M$ in $M^o$ based at $m_1$, and if we analytically continue $y'$ along this map according to the path $\gamma$, if $y'$ was in the same irreducible component of $\widetilde{C}$ as $y$ at the beginning of the analytic continuation, then the conjugate point $y''$ of $y'$ must be in the same irreducible component of $\widetilde{C}$ as $y$ at the end of the analytic continuation. Therefore, every pair of $N$-conjugate points of $(f^o)^{-1}(x_1)$ are in the same irreducible component of $\widetilde{C}$.

Finally, since there were at most $\mu$ distinct orbits of $N$ acting on $(f^o)^{-1}(x_1)$, it follows that $\widetilde{C}$ has at most $\mu$ distinct irreducible components. For each irreducible component $\widetilde{C}_i$, denoting by $d_i$ the degree of that component over $\text{Image}(u_{m_1})$, denoting by $g_i$ the arithmetic genus of (the normalization of) that component, and denoting by $b_i$ the branch number of that component, then Riemann-Hurwitz gives
$$
2g_i-2 = d_i(-2) + b_i, \ \ b_i = 2d_i - 2 + 2g_i \geq 2d_i - 2.
$$
Summing up over all irreducible components, we have,
$$
\langle [D_f], \beta \rangle \geq 2d - 2\mu \geq 2d - 2(m!).
$$

**Edit.** I realize now that there is no need to pass from $\pi_1(M^o,m_1)$ to the normal subgroup $N$. So the correct bound is not $m!$, but $m$. Also, I actually can think of no example [added: in characteristic zero] where $\widetilde{C}$ is disconnected. So it might be that the correct inequality is $$\langle [D_f], \beta \rangle \geq 2d-2.$$ [Added later:] The argument above seems to be valid even in positive characteristic (after excising all mentions of "loops"), even when $X$ is rationally connected but not separably rationally connected. Such $X$ may have finite, nonzero $\pi_1(X,x_1)$. Thus, it is probably impossible to improve much on the inequality $\geq 2d - 2m$ in positive characteristic. As far as I know, the true inequality in characteristic zero may be $\geq 2d-2$.