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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 where $\widetilde{C}$ is disconnected. So it might be that the correct inequality is $\langle [D_f], \beta \rangle \geq 2d-2$.

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 $\geq 2d-2$.

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 where $\widetilde{C}$ is disconnected. So it might be that the correct inequality is $\langle [D_f], \beta \rangle \geq 2d-2$.

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 $\text{ev}|_M$. The claim is, for every morphism $f:Y\to X$ as above, $$\langle [D_f], \beta \rangle \geq 2d - 2(m!).$$

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 $y_2$, $y_3$ of points of $(f^o)^{-1}(x_1)$ that are in the same orbit of $N$, there is a real continuous path in $Y^o$ from $y_2$ to $y_3$ whose image under $f^o$ is a loop in $X^o$ based at $x_1$ and whose homotopy class is in $N$. In particular, the loop lifts to a loop in $M^o$ based at $m_1$.

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 a loop in $M^o$ based at $m_1$, and if we analytically continue along this map, 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 $\text{Image}(u)$, 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, $$ \langle [D_f], \beta \rangle \geq 2d - 2\mu \geq 2d - 2(m!). $$

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 $$\langle [D_f], \beta \rangle \geq 2d - 2(m!).$$

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 $m_1$.

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, $$ \langle [D_f], \beta \rangle \geq 2d - 2\mu \geq 2d - 2(m!). $$