This is a simple and clear issue, with a unique answer. I see other replies mentioning weather conditions, dark adaptation and so on. That's just so much hand waving, given that the first thing you said was "I've always lived in somewhat large cities".
The core problem here, by a very wide margin, is light pollution if you live in a large city. This is the one factor, above everything else, that affects your ability to see the stars.
Here's a light pollution map:
http://www.jshine.net/astronomy/dark_sky/
The white zones are the worst, and they are in the middle of the cities. Black zones are the best.
Here's a somewhat better (but not perfect) comparison of a dark sky versus light polluted sky (your picture was taken with a very long exposure that doesn't look very realistic):
The dome of light above the city is very visible if imaged from afar:
Long exposure pictures in cities will reveal the orange skyglow, which is the main reason why you can't see the stars - it's like noise masking off the faint light from the distant objects:
Light pollution affects primarily the observations of faint objects, such as nebulae or distant galaxies. Bright objects such as the Moon, the big planets, or some of the bright stars, are not affected by light pollution.
Using a telescope with a large aperture alleviates the effects of light pollution to some extent, but it cannot work miracles. A dark sky is always better.
Usually a 1 hour drive away from the city will bring you in a place with dark sky, free of light pollution - but it depends on several factors. In such a place you should be able to see the Milky Way with your naked eye. The Andromeda galaxy also is visible with the naked eye if the sky is dark enough.
The effect that you're describing is extremely small. Have a look at the following figure:
Here you see the position of the Sun from a location $L$ on Earth. Let's call $R_\oplus$ the radius of the Earth, $\Delta$ the distance between the Earth and the Sun, and $\varepsilon$ the obliquity of the Earth. The angle $\delta$ is the declination of the Sun at a particular day, i.e. the angle between the Sun and the equator. $\delta$ varies between $-\varepsilon$ (at the winter solstice) and $\varepsilon$ (at the summer solstice), with $\delta=0^\circ$ at the equinoxes. More specifically,
$$
\sin\delta = \sin\varepsilon\sin\lambda,
$$
where $\lambda$ is the ecliptic longitude of the Sun at that day.
If $\varphi$ is the latitude of the location, then
$$
\theta = \varphi - \delta
$$
at the given day. Now suppose that the obliquity changes from $\varepsilon$ to $\varepsilon'$. Then for the same location and the same day, the solid lines change into the dashed lines, with
$$
\begin{align}
\sin\delta' &= \sin\varepsilon'\sin\lambda,\\
\theta' &= \varphi - \delta'.
\end{align}
$$
The change in position of the Sun, with respect to the distant stars, will be
$$
p = \alpha' - \alpha.
$$
These angles can be calculated with basic trigonometry:
$$
\begin{align}
\ell\sin\alpha &= R_\oplus\sin\theta,\\
\ell^2 &= R_\oplus^2 + \Delta^2 - 2R_\oplus\Delta\cos\theta,
\end{align}
$$
thus
$$
\sin\alpha = \frac{R_\oplus\sin(\varphi - \delta)}{\sqrt{R_\oplus^2 + \Delta^2 - 2R_\oplus\Delta\cos(\varphi - \delta)}},
$$
and analogously
$$
\sin\alpha' = \frac{R_\oplus\sin(\varphi - \delta')}{\sqrt{R_\oplus^2 + \Delta^2 - 2R_\oplus\Delta\cos(\varphi - \delta')}}.
$$
The dominant term in the denominator is $\Delta^2$, and it is clear that both $\alpha$ and $\alpha'$ are small angles; since $\sin x\approx x$ for small $x$ (in radians), we get
$$
p^\text{(rad)}\approx \frac{R_\oplus}{\Delta}\!\left[\sin(\varphi - \delta') - \sin(\varphi - \delta)\right].
$$
Since $R_\oplus/\Delta\approx 0.000043$, the resulting parallax is extremely small.
Edit
The value of $p$ will be maximal when
- the change in $\delta$ is maximal, i.e. at the solstices. At the summer solstice, we have $\delta=\varepsilon$ and $\delta'=\varepsilon'$.
- the angles $\varphi-\delta$ and $\varphi-\delta'$ are minimal (because the change in the sine function is maximal for small angles). For instance, we can take the latitude $\varphi=\delta$.
With these assumptions, we get
$$
p^\text{(rad)}\approx \frac{R_\oplus}{\Delta}\!\left[\sin(\varepsilon - \varepsilon')\right].
$$
If we take $\varepsilon=23^\circ$ and $\varepsilon'=24^\circ$, then
$$
\begin{align}
p^\text{(rad)}&\approx -\frac{R_\oplus}{\Delta}\!\sin(1^\circ)\\
&\approx 7.5\times 10^{-7},
\end{align}
$$
which corresponds with $0.15''$, i.e. about a tenth of an arcsecond. For comparison, let's look at the angular size of Orion:
In equatorial coordinates, its borders lie between about $4^\text{h}43^\text{m}$ and $6^\text{h}25^\text{m}$ in right ascension, and between $23^\circ$ and $-11^\circ$ in declination, which is about $25^\circ\times 33^\circ$. In other words, Orion is many orders of magnitude larger than the change in solar position.
Best Answer
According to this article:
So a human observer should be able to see roughly half of the visible sky, or a quarter of a full sphere. If you look about 50° above the horizon, you should be able to see an area of the sky extending from directly to your left to directly to your right horizontally, and from the horizon to about 10° past the zenith.
You won't be able to see as well around the edges of your field of view. Depth perception isn't relevant when looking at the sky, but being able to see with both eyes is probably better than seeing with just one.
This can vary from person to person; some people have much better peripheral vision than others.