Given an irreducible curve $C$ of degree $d$ in $\mathbb{P}^r$ and a general hyperplane $H\subset\mathbb{P}^r$, the uniform position theorem states that any $r$ points on the hyperplane section $H\cap C$ will be linearly independent. (Suppose we are working over the complex numbers)
One reference I have found so far is Arbarello-Cornalba-Griffiths-Harris, Geometry of Algebraic Curves I.


Now I wonder, what general really means in this statement:


My first hope is that any transversal intersection will do. Is this true? My feeling is that this might not be good enough.


Should it fail, suppose I am given a subspace of linear forms on $\mathbb{P}^r$ that induces a base point free linear system on the given curve $C$. Can I find a general hyperplane section in this linear system such that the uniform position property holds?



Uniform position property and general hyperplanes
The answer to both questions is negative, as pointed out by aginesky in the comments. To complete his answer, let me give a simple concrete counterexample.


Consider a canonical curve $C$ of genus 4 with only one $g^1\_3$, so a complete intersection of an irreducible singular quadric $Q$ and a cubic $G$ in ${\mathbb P}^3$. 


Pick a line $l$ through the node of the quadric which is NOT contained in $Q$, and consider the pencil of planes containing $l$. Since $l$ intersects $Q$ just in the node, which is not in $C$, the pencil has no base points on $C$, as you require.


On the other hand, the intersection of each of those planes with $Q$ splits as two lines, and therefore gives a divisor of degree 6 on $C$ which fails uniform position: more precisely the 6 points split as two sets of three collinear points.


