Why does the homotopy definition represent the usual drawings? When motivating the definition of a homotopy, one often sees the picture of the image of two paths being "connected" by the images of other paths. I am a bit irritated as to why the definition of a homotopy captures this intuition, since the domain as well as the "movement" of the functions, is rarely depicted in these motivations. What these pictures do is they deform the image of the function, not the function in a continuous way, don't they? Therefore I wondered whether these pictures accurately motivate the definition of a homotopy or why homotopy isn't defined in a way that captures the deformation of the images.
I also wondered whether this can lead to false intuitions as follows. Let $I$ be the unit interval and $\mathrm{id}$ be the identity on $I$. Furthermore let $$f:I \to I, \begin{cases}
x \mapsto 2x,  \ x \leq \frac{1}{2} \\
x \mapsto 1, \  \  \ x > \frac{1}{2}
\end{cases}$$
Now defining $H:I \times I \to I$ to be the identity at time $0$ and $f$ at all other times, we get the an image of how a homotopy is motivated, but this doesn't define a homotopy, since $t \mapsto H(\frac{3}{4},t)$ is not continuous at $t=0$.
When drawing the usual picture however, one might naively say that they are homotopic, since one can deform one image into the other continuously.
Are there "better" ways to motivate the definition of a homotopy? I think one way is to include the domain of the functions and to indicate how the domain is mapped. This would at least prevent the example I mentioned above and may make the intuition more precise.
 A: Yes, drawing the usual picture suggests that $f\simeq 0$. That doesn't mean that any function $H:I\times I\to I$ will do, it just 'means' that there is, or should be, a valid $H$. Your $H$ doesn't work because it jolts straight from $0$ to $f$ - it is not a continuous deformation, both in the rigorous and informal sense.
A correct $H$ might be: $$H:(x,t)\mapsto\begin{cases}2xt&0\le x\le\frac{1}{2}\\t&\frac{1}{2}<x\le 1\end{cases}$$
This provides $0\simeq f$ and it does so exactly as intuition might expect. The visual intuition is very important because it allows me, and mathematicians generally, to think concretely about 'what's going on': it's easy to get lost in symbols and formalism, and accordingly get very stuck! Of course, when the time comes to prove things, we should be more precise, but there's nothing wrong with intuitive scratchwork.

I drew this image in my head, and was able to figure out a suitable $H$ very quickly. The axes should be interpreted loosely. I don't claim that my $H$ is the only valid $H$, of course.
As for your bolded questions at the start, I would say that deforming the image of a function is exactly the same as deforming the function. Since neither term is precisely defined, that's ok for me to say, and both terms really should mean the same thing: "changing (continuously) the values of a function at a point" applies equally to both.
One reason why homotopy is often talked about in such terms is the path-space adjunction: $$C(X\times I,Y)\cong C(X,Y^I)$$
'Naturally' in $X$ and $Y$. Here, $Y^I$ is the set of all continuous functions $I\to Y$ as topologised by the compact-open topology. This exponential object behaves as we (categorically) wish because $I$ is strongly locally compact Hausdorff. What this is saying, is that homotopies of maps $X\to Y$ are in 'natural' correspondence with maps $X$ to the path space of $Y$, which is the assignment - to every $x\in X$ - of a path in $Y$. You can think of this, insofar as it relates to homotopy, as the path $H(x,t)$ in $t\in I$ for a given homotopy $H$ - the continuous deformation of the values $f(x)$ to $g(x)$. In my image, these are the blue lines.
