Two vectors (starting or shifted/imagined to start at the same point) always belong to the same plane, usually one plane. If the vectors are $A\to B$ and $A\to C$, just imagine that you connect $B,C$ by a straight line, thus completing a triangle, $ABC$. This triangle is already a clear "seed" of a plane, isn't it? The plane isn't necessary vertical, horizontal, or parallel to any other plane you may have thought about at the beginning but it is a plane nevertheless. There always exists a vector $\vec n$ that is orthogonal to the whole plane or, equivalently, that is orthogonal both to the vectors $A\to B$ and $A\to C$.
This is not really physics but rather geometry for the 10-year-olds.
Let's go back and consider the Biot-Savart Law for a filamentary current:
$$B(r) \propto \int_C I(r') \times \frac{r-r'}{4\pi |r-r'|^3} \, dr'$$
As I've written it, $r'$ is some position on the coil (we integrate over the coil to get the whole magnetic field). $r$ is the position we want to find the field at--somewhere inside the coil's empty body.
Let's consider $r$ directly at the center of the coil. As we start integrating, we slide along the coil. I'll traverse the loop clockwise, so let's have $r'$ start at the top of the batter, come across the top of the page, and start traversing the topmost loop of the coil.
As we start on the rightmost part of the topmost loop, the instantaneous direction of current is out of the page. If we take $r = 0$, so that the center of the loop is the origin, then all we have is the vector $I(r') \times [-r'/|r'|^3]$ as the integrand. $-r'$ points inward toward the center of the coil. It should be clear that the resulting magnetic field from a small piece of the wire at this point in the coil is both downward and to the left.
Let's consider what happens when we get to the leftmost part of the topmost loop. The vector $r'$ is downward and to the left. The current is into the page. The resulting magnetic field is downward and to the right.
In general, as we traverse the topmost loop, each small piece of wire adds a magnetic field that is (a) downward and (b) pointing out of the coil. (I must remind that we're talking about the magnetic field only at the center of the whole coil right now).
If the topmost loop were rotationally symmetric, we could argue that any components that point away from the central axis of the coil must cancel. The real coil does not have this symmetry, but it's "pretty close" to being rotationally symmetric, and any such real components ought to be small.
All the other loops work basically the same way, contributing only net downward magnetic field when integrated over a whole circular loop.
For some reason, you referred to the magnetic field contribution from different points as being clockwise or anticlockwise. I do not understand this. This coil does not create any kind of closed magnetic loops that can be seen on this scale. The field is more clearly described using fixed directions (into or out of the page, left or right, down or up).
I think it's this reason that you thought "clockwise and anticlockwise should cancel". But you described it more correctly in saying that at both points, there is a downward component; the only thing that can cancel a downward component is an upward component! You were right to say that at points where the current moves left, the field's direction is down and out of the page, and that when the the current moves right the field direction is down and into the page. It's just that only the into/out of page components cancel, and net downward is left behind.
Best Answer
A nice bubble chamber picture:
"Electrons and positrons produced simultaneously from individual gamma rays curl in opposite directions in the magnetic field of a bubble chamber. In the above example the gamma ray has lost some energy to an atomic electron, which leaves the long track, curling left. The gamma rays do not leave tracks in the chamber, as they have no electric charge"
Hans.