I don't know what's wrong with using a white card to find the focus. The focus is where you see the image of the lamp filament. You cannot focus the filament to a smaller image size than the actual filament itself without running up against the laws of thermodynamics.
EDIT: OK, this answer is really wrong (see comments below). Leaving it up because I don't believe in covering my tracks.
EDIT 2: I'm going to try this one more time. What is probably true is that no system of lenses is going to cram more light into the fibre than you would get just by butting up the end of the fibre directly against the lamp. Any takers?
EDIT 3: Second answer also wrong. Let's try this: Express the capture area of the lens as a fraction of the total emission angle of the source. That is the same fraction which gives you the minimum (optimum) image size as compared to the size of the source.
(Also, re-reading the original question, I think I now understand the point about image point versus focal point: the focal point is what you get with parallel rays, i.e. a source at infinity. If the source is close, so you don't have parallel rays, the focus moves correspondingly. The focus is literally where you see the image on a white paper, and the calculated focal length of 220 mm is irrelevant.)
EDIT 4: My third edit assumes that your source and target are in fixed locations, and you have an infinite number of lenses with different focal lengths to play with. I will now, finally I hope, try to address the actual optimization of the system which you actually have: the big Fresnel lens with the 220-mm focal length. First let's imagine the "symmetric" case, where the source and the target are 880 mm apart, and the lens is right in the middle. When you put your white piece of paper at the target location, you should see an image size exactly the size of the actual filament. You can see this should be so because of the symmetry of the ray tracing patterns. Let us examine what happens if you try longer or shorter combinations.
First imagine what happens if you put your eyeball directly at the image point and look at the lens. (Imagine your eyeball to be the size of a geometric point.) If your eyeball is placed within the actual filament image, then when you look at the lens you will see the entire surace illumainated with the fiery color of a hot filament; if your eyeball moves outside the image zone, you see merely ambient light, or darkness as it may be. Now let us move our target backwards, away from the lens. Which way do we have to move the lens to adjust the image to be on target? Answer: if we move the lens in the opposite direction, towards the target, we will find a new focus. It is easy to see this because if we move our target all the way back to infinity, the lens must be 220-mm away from the source. (Remember we said we were starting at optimum conditions with the lens 440-mm from the source.) If we put our eyeball exactly right on target, we still see the lens filled with that fiery color; but we are so far away that the lens looks very small, and therefore the total amount of light it delivers is small. The image is actually quite large, but very faint.
But wait: if instead of moving the lens towards the source, if we had instead followed the target, we would also find a solution at infinity where the lens is 220 mm from the image point and the light bulb is at infinity! If we put our eyeball right at the image point, we once again see the bright fiery color of the hot filament; but this time, it does not fill the whole size of the lens, which is close to us and looks quite large. The image is now quite bright, (like when you focus sunlight) but it is absolutely tiny, so if we move our eyeball off target in the slightest, we see merely ambient light, or darkness. Again, the total amount of light collected is small.
To summarize: the closest proximity of target to source at which the lens focuses an image is 4x the focal length; at this position, the image size is the same as the actual size of the source. There are no closer solutions. For every longer case of source-to-target distance, there are two solutions for lens position; in one case, the image is larger and fainter, and in the other case, the image is smaller and brighter.
Where is your maximum light collection? As you bring the lens closer to the source, you are intercepting a bigger solid angle, so you are collecting more light; but the image size at the focus is also getting bigger (and fainter), so eventually your are not able to concentrate all the availble light into your target. I'm guessing the maximum light collection occurs when your image size is the same diameter as your optical fiber, because that's the closest lens-to-source proximity and hence the largest solid-angle interception without "spillage" at the point of collection.