In the lecture course that I'm currently taking on Fourier transforms, we defined the Fourier transform of a function $f:\mathbb{R}\to\mathbb{R}$ as $$\hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx} \, dx.$$ For this to be well defined we require that $f$ be absolutely integrable and have only a finite number of discontinuities. The absolute integrability condition implies the $f$ decays at $\pm \infty.$
We later defined the Fourier cosine transform for a function $g:(0,\infty)\to \mathbb{R}$ as $$\hat{g}(k)=\int_0^{\infty}g(x)\cos(kx)\, dx.$$
I'm unsure of what we require of $g$ for this to be well defined. I think we need that $g$ decays at $+\infty$ and has only a finite number of discontinuities, but I'm not sure if we need any other conditions (e.g., $g$ decays at $0$).
It does not. For example, $f$ could be a piecewise linear function whose graph contains a triangle with vertices $(n, 1)$, $(n\pm 1/n^2, 0)$. This is a typical example of an absolutely integrable function that does not tend to zero at infinity.
Same as for the original form. The cosine is just the real part of complex exponential. There is no particular reason for the former to converge if the latter does not.
I'll note that the conditions you quoted were chosen by the lecturer for the convenience of presentation; Fourier transform can be defined in greater generality. For example, "has finitely many discontinuities" could be replaced by "is measurable".