Say $f:X\to C$ is a family of curves. More precisely, $C$ is a smooth projective irreducible curve over a field, $f$ is a flat morphism of schemes and $X$ is a normal projective irreducible surface.
Say I take a section $P:C\to X$. Does the image of $P$ lie in the nonsingular part of $X$?
What conditions (weaker than regularity) on $X$ do I need? Does it help if every Weil divisor on $X$ is $\mathbf{Q}$-Cartier?
Example: Let $Y$ be the projective cone over a conic, so for instance, let $X$ be defined by $xz=y^2$ in the projective 3-space with coordinates $[x:y:z:w]$ and consider the projection to the $[x:w]$-axis: $$ g:Y\dashrightarrow \mathbb P^1$$ $$ [x:y:z:w]\mapsto [x:w]\quad\qquad $$ This is defined everywhere except at $[0:0:1:0]$, so let $X$ be the surface one obtains by resolving the indeterminacies of $g$ and $f:X\to \mathbb P^1$ the induced morphism.
This $X$ is a normal, projective, irreducible surface and $f$ is a dominant morphism onto $C=\mathbb P^1$. It follows that $f$ is flat. $X$ is also non-singular everywhere except at $[0:0:0:1]$ where it has an $A_1$-singularity which is one of the simplest surface singularities. In particular, it is $\mathbb Q$-factorial, i.e., every Weil divisor is $\mathbb Q$-Cartier.
There is also a section that goes through the singular point: $\sigma: \mathbb P^1\to X$ defined by $[x:w]\mapsto [xw^2:x^2w:x^3:w^3]$ which originally maps to $Y$, but lifts to $X$ by taking the strict transform of the image. Notice that the image of this section is $\mathbb Q$-Cartier.
Analysis: The above example shows that unfortunately what you want does not necessarily hold, even under pretty strict conditions. The only condition short of non-singularity I could imagine is to assume that $X$ is factorial, that is, every Weil divisor (or at least the section you are considering) is Cartier. This is of course very easy: the section is non-singular since it is isomorphic to $C$ and if a variety contains a smooth Cartier divisor, then it is smooth along that divisor.
One could prove this last statement in a little more complicated way that shows why this fails in the example above:
The intersection of the section and any fiber has to be $1$, because the projection is an isomorphism. If the section is Cartier, this means that the fiber cannot be singular at the intersection point, so not only $X$ is non-singular, but even $f$ is smooth along the section. The problem with this argument in the case when the section is only $\mathbb Q$-Cartier is that the intersection number of the section with a fiber could be $1$ and the fiber still be singular.
In the example, the section is a smooth curve, but it is not Cartier (this actually follows from the above, but it is usually proved by computing the self-intersection, which is $\frac 12$), but the fiber through the singular point of $X$ is singular and the section goes through that point. Twice the section is Cartier and its intersection with the singular fiber is $2$, so the intersection of the section with the fiber is $1$ and we get no contradiction.
