No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{O}(e_2))$$ If $e_1$ and $e_2$ were both positive, then term on the right would be at least $4$. So this is impossible.

No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{O}(e_2))$$ If $e_1$ and $e_2$ were both positive, then term on the right would be at least $4$. So this is impossible.

Added in response to comment. If you are allowed to pick $\deg L\gg 0$ relative to $k$, then I think it's probably true. Here's a result in that direction.

Lemma. If $\deg L\gg 0$ relative to $k$, then all $e_i\ge 0$.

Sketch. We can assume $L=\omega_{C/\mathbb{P}^1}(M)$ with $M$ globally generated. By a standard trick, we can find a cyclic cover $\pi:\tilde C\to C$ such that $L$ is a direct summand of $\pi_*\omega_{\tilde C/\mathbb{P}^1}$. Then $p_*L$ is a summand of $(p\circ \pi)_*\omega_{\tilde C/\mathbb{P}^1}$. The last sheaf is semipositive by a theorem of Fujita.

I suspect with more work, you can make the $e_i$ positive, but I leave that to you.

No, not in general. Take $C=\mathbb{P}^1$, $L=O(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but $h^0(p_*L)=2$, so $e_1$ and $e_2$ can't both be positive.

No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{O}(e_2))$$ If $e_1$ and $e_2$ were both positive, then term on the right would be at least $4$. So this is impossible.