I will sketch a counterexample for the modified question. The idea behind the construction is similar to the counterexample for the original question. Only the geometric details of this construction are trickier and does not integrate so well in my previous answer. For this reason I will post this a new answer.

For this example we take $k=\mathbb{C}$ and $C=\mathbb{P}^1$.
Take two polynomials $a(t), b(t)$ of degree 2 and 4 and consider the elliptic  curve $E_{a,b}: y^2=x(x^2+a(t)x+b(t))$. This is a rational elliptic surface, with Mordell-Weil group $\mathbb{Z}^4 \times \mathbb{Z}/2\mathbb{Z}$, provided that $a$ and $b$ are sufficiently general, e.g., $a$ and $b$ have no common zero and both $b$ and $a^2-4b$ are squarefree.

Take a point $Q\in E_{a,b}(K(t))$ of infinite order. Let $T$ be a point of order 2, different from $(0,0)$ and take for $P$ the point $Q+T$. Then $2P\in E_{a,b}(\mathbb{C}(t))$ and if $a^2-4b$ is not a square then $P$ is not in $E_{a,b}(\mathbb{C}(t))$.

As in my previous answer, it suffices to find a $(a,b)$ such that the twist of $E_{a,b}$ over $K(P)=K(\sqrt{a^2-4b})$ has positive rank. I will sketch two arguments why such an $(a,b)$ exists.

Let $U \subset \mathbb{C}[t]_2\times \mathbb{C}[t]_4$ be the locus of polynomials such that $E_{a,b}$ defines a rational elliptic surface with positive Mordell-Weil rank and precisely one point of order $2$. (The complement of $U$ is the union of the locus where $a$ is a multiple of $b^2$, together with the locus where $a^2-4b$ is a square, and finitely many codimension 4 components parametrizing the Mordell-Weil rank $0$ surfaces.)

For any integer $m$ consider the locus $L_m$ of $(a,b)$ such that the twisted elliptic curve $(a^2-4b)y^2=x(x^2+ax+b)$ has a section of infinite order which intersects the zero section in $m$ points. 
Using Hodge theory, Lefschetz $(1,1)$ and the fact that the corresponding elliptic surface has $h^{2,0}=2$ it follows that this locus is locally given by two equations. This implies that $L_m$ is either contained in the complement of $U$ or $L_m$ contains a component of codimenion at most two in $U$.
Now the complement of $U$ has only one component of codimension two. Since for most $m,m'$ we have that $L_m\neq L_{m'}$ we find at least one component of some $L_m$ which is not contained in $U$. Actually, there is a standard argument in Noether-Lefschetz theory to check whether the union of $L_m$ is dense in the analytic topology on $U$ and I expect that this argument applies also here.

A second way to construct an explicit example. It is rather easy to find equations for the loci $L_m$. For $L_0$ you have $19$ equations in $25$ variables. Since the equations are of small degree I expect that with some help from a computer you can find an explicit example. Moreover, modern computer algebra software should be able to check whether $L_0$ is in the complement of $U$ or not.


Edit: If you want that $E_{a,b}$ is semistable then you need to do a bit more work. In this case you need to exclude all $(a,b)$ with a common zero from $U$. This defines a codimension one locus. Hence, depending on the strategy you need to show that $\cup L_m$ is dense or that $L_0$ is not contained in the complement of $U$.