When it comes to light any analogy with water is doomed to be completely wrong.
I word this so strongly because it is a common misconception that frustrated all of physics for several decades back around the turn of the 20th century. People kept trying to make analogies with water, and the result was always wrong.
What you are describing is what is known as the aether theory of light. People thought "well, light is a wave, and water waves need a medium to travel through, therefore there must be some sort of medium in which light travels." This is simply not true. The EM field is not some sort of physical substance that has a bulk velocity. When light is emitted, it is not perturbing some medium that is stationary in some reference frame. It propagates, and you can describe that propagation in any frame you like, but there is no "correct" or "natural" frame.
Michelson and Morley are most famous for designing the experiment that proves this beyond any reasonable doubt. Basically, they measured the speed of light in various directions, aligned with and perpendicular to the Earth's motion about the Sun. That speed did not change, not one bit. The Sun's rest frame is nothing special; the source may be moving with respect to it, but so what? Light moves away from its source, at the speed of light, in a "straight line," as seen by any inertial observer.
If you thought otherwise, you would have to believe that some inertial frames are more important than others; that some observers/sources are really, truly, absolutely in motion while others are truly at rest; that the shore of the pond was a more privileged thing in the universe than person running alongside it. Physicists eventually realized this was preposterous - the universe doesn't think the shore of the pond is more special than a person - and hence was born special relativity.
Yes, you will see the light beam traveling in a straight line. The principle of relativity states that all laws of physics are the same for observers in relative inertial motion. Therefore, you should see no difference in how light propagates whether you are standing still or moving rapidly, as long as you are moving rapidly with a constant velocity. However, in special relativity you cannot simply add velocities the way you are used to from high school physics. Otherwise there would be nothing preventing you from breaking the "cosmic speed limit" of the speed of light. Any time you walk while holding a flashlight you would be producing light moving at a speed $c +$ your walking pace! If you add colinear velocities $v$ and $u$, the result will be $$\frac{v+u}{{1+\frac{vu}{c^2}}}$$
You can see that if you consider velocities much lower than $c$, the denominator will approach $1$, and you recover the familiar formula $v+u$.
If you fire the laser off at an angle from the direction in which you are moving, then there is an additional consideration you must take into account. Angles will also change depending on your reference frame. This is behind a phenomenon known as relativistic beaming. If you were to take a light source that radiated in all directions, like a light bulb, and suddenly set it moving at high speed, there would be a higher intensity of light emitted along the direction it is moving as compared to other directions. The reason for this is that when a light ray is emitted along the angle $\alpha$ from the light bulb's perspective, it is emitted at the angle $$\alpha' = \cos^{-1}\left(\frac{\cos\alpha + V}{1+V\cos\alpha}\right)$$ from your perspective.
The last part of your question is a little difficult for me to decipher, but I hope this has been of some help.
Best Answer
It's not that light has any inertia, it's simply that when you view the path of something from a different frame, the path can look different. Light's having no mass does not affect this qualitative fact. The masslessness simply means that its speed is frame-invariant, not its direction of travel.
In particular, if for example, a light ray is traveling in the $y$-direction in one frame, then it will look as though its path is angled towards the negative $x$-axis in a frame boosted along the positive $x$ axis.
Mathematically, consider a light ray along the $y$-axis described by the following path: \begin{align} t(\lambda) &= \lambda, \qquad x(\lambda) = 0, \qquad y(\lambda) = c\lambda \qquad z(\lambda) = 0 \end{align} The Lorentz transformation tells us that in a frame boosted in the $x$-direction, we have the following relationships: \begin{align} t' &= \gamma\left(t-\frac{v}{c^2} x\right), \qquad x' = \gamma(x-vt), \qquad y'=y, \qquad z'=z \end{align} so the path of the light ray in the new frame is \begin{align} t'(\lambda) = \gamma\lambda, \qquad x'(\lambda) = -v\gamma\lambda, \qquad y'(\lambda) = c\lambda, \qquad z'(\lambda) = 0 \end{align} Notice that the path of the ray has picked up a negative $x$ compoenent! This is purely because we are viewing the light ray in a different way. It's like if you were to view your computer monitor while sitting at your desk, it looks like its standing still, but you can make it look like its moving in any direction you please by yourself moving in an appropriate direction.
Notice, however, that in both frames, the path of the light ray satisfies the appropriate nullness condition; \begin{align} -c^2\dot t(\lambda)^2 +\dot{\mathbf x}(\lambda)^2 = 0 \end{align} where dots denote derivatives with respect to the parameter $\lambda$ which indicates that the speed of the light ray is invariant.