How can an aircraft turn if the horizontal force component is zero?
It is easier if we look only at the forces experienced by the aircraft, and in an inertial frame of reference



In this revised diagram, the vertical component of the lift balances the weight, which is vertical. There is a remaining horizontal component of the lift, and this causes the turning.

"Centrifugal force" does not exist (and is not needed) in an inertial frame of reference 

The problem with the original diagram in the question is it superimposes an imaginary force, the centrifugal force, upon the real list of forces on the aircraft.

It is hard for the general public to understand Newton's first law of motion, that any object tends to travel in a straight line when no force is acting it. It is hard for them to understand that motion in a circle is dramatically different from straight-line, constant-speed motion, since both seem (in a sense) steady or continuous.

"Centrifugal force" is a term produced by humans to describe what they think must be happening

In the case of a passenger going round a corner in the car, this inward force is applied by the outer door of the car, against which the passenger ends up leaning during the turn. It is an inward (centripetal) force, but for the general public who is not aware of the special status of inertial frames of reference will likely use the car (which seems large and solid and therefore plausible) as a frame of reference and say "something must be pushing me against the door" rather than "the door is pushing me towards the centre of the circle". 

Because of this use of a non-inertial frame of reference, the general public explanation needs to an extra force to explain why objects seem to accelerate towards the outer door of the car and the passenger feels pressed against it. In an inertial frame of reference it is the outer door of the car pressing in on the passenger, and accelerating towards any objects moving freely inside the car.

In everyday life we sometimes use special frames of reference for convenience

We say the sun "comes up" in the morning, rather than the earth "rotates bringing the sun into view". This is because it is convenient for humans to use a non-inertial, but readily identifiable, frame of reference. It does not mean that people really think that the sun orbits around a stationary earth.

It is perfectly reasonable to imagine the sun travelling around the earth, for convenience, but if we want to make calculations about space travel with multiple objects (stars, planets etc) affecting a body's motion, it will get very complicated indeed to add in all the numerous centrifugal forces etc we would have to add in. It would be simpler to use an inertial frame of reference, and all objects moving relatively simply in relation to that.

Likewise we can learn to be sophisticated about circular motion. Let the general public speak of "centrifugal force" as this simplifies conversation, but when trying to calculate things, use an inertial frame of reference to avoid agony.

Even apparently inertial frames of reference may have unconsidered rotation  in relation to a larger-scale environment

Jan Hudec in the comments below brilliantly deflates the above criticism of rotating frames of reference by pointing out that even my notionally inertial frame of reference (the earth) is in fact rotating, so that the gravitational force measured is lower than it would be for a plane moving similarly above a non-rotating earth, i.e. is attenuated by a centrifugal component. Ouch!

And of course the earth is rotating around the sun, and the sun around the galaxy, etc: 

Big fleas have little fleas,
Upon their backs to bite 'em,
And little fleas have lesser fleas,
and so, ad infinitum.


So it is just a question of how much rotation one wants to accept as part of one's frame of reference. My suggestion is that when a questioner is feeling puzzled as the original poster was, a good way to break the problem down is to keep the frame of reference outside the rotation being studied, so that the studied rotation appears as a rotation: in this case the only acceleration (and therefore force on the orbiting object) will be centripetal.