Time and speed of light in Relativity Time running slower near a massive object, but the speed of light does not really change near a massive object, according to Relativity - it just curves. Is not time directly related to the speed of light in all time-related equations in the Theory of relativity, or I get it somehow wrong? The way I understand it s that time is always connected and relates to the speed of light. I am a bit confused about this and will be grateful to get some help!
 A: Don't think of time as a "thing" that "moves" or "runs". Think of it as just another direction in which we can move. People in different places in spacetime will have different definitions of how that movement splits into space and time.
As an analogy, suppose Alice and Bob are standing together at the equator, and Alice starts moving northwards at a constant speed. For Alice, she is always moving straight ahead (in the same direction) and at the same speed, but for Bob the amount by which she's moving northwards decreases, and indeed  when she reaches the North Pole her northwards velocity is 0.
Similarly, a clock near a massive object always moves forward in its time direction at 1 tick per second, but a distant clock in "empty space" will, because of the curvature of spacetime, view the clock as not moving as fast. That's not because the clocks are different, it's because their "time directions" are pointing differently.
A: from your comment to Eric's fine answer, I think you do not really understand how the second is defined.
Please refer to the relevant Wiki page.
In particular....

When the atom is irradiated with electromagnetic radiation having an
energy corresponding to the energetic difference between the two
sub-levels the radiation is absorbed and the atom is excited, going
from the F = 3 sub-level to the F = 4 one. After a small fraction of a
second the atom will re-emit the radiation and return to its F = 3
ground state. From the definition of the second it follows that the
radiation in question has a frequency of exactly 9.19263177 GHz,
corresponding to a wavelength of about 3.26 cm and therefore belonging
to the microwave range.

So an exited atom will emit a EM radiation at a very well defined frequency.
An observer with this atom sitting in a gravitational well (Earth, for example) with a receiver and an oscilloscope, will see a wiggling voltage on her screen. When she sees  9.19...billions crossings of the zero line in the up direction she says : "a second has elapsed!" according to her atomic clock.
However the same signal will proceed to outer space...far away in the almost flat asymptotic region. Another observer up there, with a receiver and oscilloscope will see the same 9.19..billions crossings of the zero line in the up direction, but... they will take more than a second by her own atomic clock.
Why is that? Well you know the gravitation red shift? Radiation originating in a gravitational well, is red-shifted (ie. has lower frequency) when received by a distant observer. So if the starting frequency was 9.19...GHz , it will be lower, say 9.10... GHz for the distant observer. So for the distant observer there will be fewer wiggles in her own "second" and 9.19...billion wiggles will take longer.
Why do we have gravitation red shift? Well Einstein reasoned the the photon , like any ascending projectile, will lose energy as it tries to leave a gravitational well. Since for a photon energy is proportional to frequency , it will have a lower frequency when arriving at the distant observer.
So, photons leaving a gravitational well lose energy, thus they "lose" frequency and since frequency is simply "number of wiggles divided by time" and the number of wiggles is the same for both observers, time has to "change" for the 2 observers.
Einstein made this reasoning when starting to understand General Relativity.
