PeroK said:
So, where do you go from here? I would claim you are new to the concept of having the same events described in two different reference frames and, in particular, cannot reconcile the invariance of ##c## with your mental picture of the motion of light. What do you do next? Accept you are wrong? Persist in your error? The choice is yours. No one can force you to understand SR.
Let's see, whether we can make the entire trouble concrete. Since it's only about kinematics, let's just consider a massless scalar field. Wo so we can simply look at the spherically symmetric solution of the wave equation,
$$\Box \Phi(x)=0, \quad \Box=1/c^2 \partial_t^2-\vec{\nabla}^2.$$
Then to have a nice model for a "light source", just consider a spherical harmonic wave, i.e.,
$$\Phi(x)=\phi(r) \exp(-\mathrm{i} \omega t), \quad r=\vec{x}.$$
The solution "outgoing" solution reads
$$\phi(r)=\frac{1}{r} \exp(\mathrm{i} k r), \quad k=\omega/c.$$
This is the solution for a light source sitting at rest at ##r=0## (the singularity of the wave).
Now it's convenient to write this in a manifestly covariant way, using the four-velocity of the light source ##(u^{\mu})=(1,0,0,0)##. Then you get
$$\Phi(x)=\frac{1}{\sqrt{(u \cdot x)^2-x \cdot x}} \exp[-\mathrm{i} k (u \cdot x-\sqrt{(u \cdot x)-x \cdot x}].$$
In a frame of reference ##\Sigma'##, where the light-source moves with velocity ##v= c \beta## in the positive 1-direction you have
$$u'=\gamma (1,\beta,0,0).$$
Then in this frame the wave is represented by
$$\Phi'(x')=\frac{1}{\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}} \exp[\mathrm{i} k [\gamma (ct'-\beta x')-\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}],$$
where I made use of ##\gamma=1/\sqrt{1-\beta^2}## to get
$$r=\sqrt{(u' \cdot x')-x' \cdot x'}=\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}.$$
As you see for an observer in this frame the singularity is at
$$\gamma^2 (x'-v t')^2+y^{\prime 2}+z^{\prime 2}=0 \; \Rightarrow \; x'=y'=0, \quad x'=v t'$$
i.e., the singularity moves with velocity ##v## in the positive ##x'##-direction along the ##x'## axis, as constructed.
At the same time the motion of the surfaces of constant phase ##0## is given by
$$\gamma (ct'-\beta x')-\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}=0,$$
and indeed after some algebra you find that it's given by
$$c^2 t^{\prime 2}=\vec{x}^{\prime 2},$$
i.e., the surface of constant phase is moving with the speed of light out from the origin.
That must be so because of the properties of the Lorentz transformation, i.e., because ##x \cdot x=x' \cdot x'##.