How does the fragment shader know what variable to use for the color of a pixel?

Furthermore, we have a predefined gl_FragColor.

Let's start with this. No, you don't have the predefined gl_FragColor. That was removed from core OpenGL 3.1 and above. Unless you're using compatibility (in which case, your 3.30 shaders should say #version 330 compatibility at the top), you should never use this.

Now, back to user-defined fragment shader outputs. But first, a quick analogy.

Remember how, in vertex shaders, you have inputs? And these inputs represent vertex attribute indices, the numbers you pass to glVertexAttribPointer and glEnableVertexAttribArray and so forth? You set up which input pulls from which attribute. In GLSL 3.30, you use this syntax:

layout(location = 2) in color;

This sets the color vertex shader input to come from attribute location 2. Before 3.30 (or without ARB_explicit_attrib_location), you would have to either set this up explicitly with glBindAttrbLocation before linking or query the program for the attribute index with glGetAttribLocation. If you don't explicitly provide an attribute location, GLSL will assign a location arbitrarily (ie: in an implementation-defined manner).

Setting it in the shader is almost always the better option.

In any case, fragment shader outputs work almost exactly the same way. Fragment shaders can write to multiple output colors, which themselves get mapped to multiple buffers in the framebuffer. Therefore, you need to indicate which output goes to which fragment output color.

This process begins with the fragment output location value. It's set very similarly to vertex shader input locations:

layout(location = 1) out secColor;

There are also the API functions glBindFragDataLocation and glGetFragDataLocation, which are analogous to glBindAttribLocation and glGetAttribLocation.

If you don't do any explicit assignments, implementations usually will assign one of your output variables to location 0. However, the OpenGL standard does not require this behavior, so you should not depend on it either.

Now to be fair, your program should have failed to link when you used two outputs that didn't get different output locations. What probably happened was that your compiler optimized the one you didn't write to out, so it kinda forgot about it when it came time to check for linker errors.

I'd like to specify this for OpenGLES 3.1 which uses GLSL_ES_3.10 link:

§4.4.2

If there is only a single output [in the fragment shader], the location does not need to be specified, in which case it defaults to zero.