Sampling Theorem and Dirac Comb
Let $X(f)$ denote the Fourier transform of $x(t)$ where
$$\begin{align*}
X(f) &amp;= \int_{-\infty}^{\infty} x(t) \exp(-j2\pi ft) \mathrm dt\\
x(t) &amp;= \int_{-\infty}^{\infty} X(f) \exp(+j2\pi ft) \mathrm df
\end{align*}$$
which I will denote via $x(t) \leftrightarrow X(f)$.
The following transform pairs will be needed in what follows.
$$\begin{align*}
\delta(t) &amp;\leftrightarrow 1\\
\delta(t-t_0) &amp;\leftrightarrow \exp(-j2\pi f \,t_0)\\
\sum_{n=-\infty}^{\infty}\delta(t-nT) &amp;\leftrightarrow
\frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\
\end{align*}$$

Given a signal $x(t)$, its sampled pulse train (at intervals of $T$ seconds) is 
$$x(t) \sum_{n=-\infty}^{\infty}\delta(t-nT) 
= \sum_{n=-\infty}^{\infty} x(t)\delta(t-nT) 
= \sum_{n=-\infty}^{\infty} x(nT)\delta(t-nT).$$
Since multiplication in the time domain corresponds to convolution in the
frequency domain, we have
$$\begin{align*}
x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) &amp;\leftrightarrow
X(f)\circledast\frac{1}{T}\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{T}\right)\\
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
X(f)\circledast\delta\left(f-\frac{k}{T}\right)\\
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
\int_{-\infty}^{\infty} X(f-w)\delta\left(w-\frac{k}{T}\right) \mathrm dw\\
x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT)
&amp;\leftrightarrow \frac{1}{T}\sum_{k=-\infty}^{\infty}
X\left(f-\frac{k}{T}\right) = \hat{X}(f)
\end{align*}$$
Thus, the Fourier transform of the impulse train formed by
sampling $x(t)$ at $T$ second intervals is $\hat{X}(f)$ which
is obtained by repeating $X(f)$ along the $f$ axis at intervals of 
$T^{-1}$ Hz and summing the result.
Furthermore, $\hat{X}(f)$ is a periodic function of the frequency
variable $f$ with period $T^{-1}$ Hz.  That is, for all $f$,
$$\hat{X}\left(f + \frac{1}{T}\right) = \hat{X}(f).$$  Note that
all of this holds regardless of what $X(f)$ is: $X(f)$ could be
nonzero for all $f$, and the result would still be valid.

Now suppose that $X(f)$ is zero if $f &lt; a$ or $f &gt; b$.  Then,
$\hat{X}(f)$, obtained by repeating $X(f)$ periodically along
the $f$ axis,  is nonzero only in the intervals
$$\ldots, \left[a-\frac{2}{T}, b-\frac{2}{T}\right], 
\left[a-\frac{1}{T}, b-\frac{1}{T}\right],
[a, b] 
\left[a+\frac{1}{T}, b+\frac{1}{T}\right], 
\left[a+\frac{2}{T}, b+\frac{2}{T}\right], 
\ldots
$$
and so if 
$$b -\frac{1}{T} &lt; a \Rightarrow b-a &lt; \frac{1}{T},$$
that is, the support $b-a$ of $X(f)$ is smaller than 
the repetition interval $T^{-1}$ Hz, then the repetitions
of $X(f)$ do not overlap.  Indeed as the OP's figures show, 
for a real-valued signal whose spectrum extends from $a = -f_0$
to $b = f_0$ (support of $X(f)$ is of length $2f_0$), sampling 
$x(t)$ at intervals of $T = (3f_0)^{-1}$ (and thus repeating $X(f)$ at
intervals of $3f_0$ Hz on the frequency axis)
leads to no overlap while sampling $x(t)$ at intervals of
$T = (1.5f_0)^{-1}$ leads to overlap of the repetitions
of $X(f)$.

Finally, the OP asks about negative frequencies and the
interpretation of the pictures which show positive frequencies
only.  For a real-valued signal, $X(f)$ has conjugate symmetry,
meaning that $X(-f) = X^*(f)$, and so specifying $X(f)$ for
positive values of $f$ suffices.  In any case, the pictures
he is looking at are of $|X(f)|$ and $|\hat{X}(f)|$ which
are even functions of $f$, and so showing only the positive
axis saves space, though it does make the pictures look a 
bit lopsided since only half the lobe is shown at low frequencies.
For complex-valued signals, $X(f)$ does not have conjugate
symmetry and $|X(f)|$ need not be an even function of $f$
and so the whole axis would need to be shown.  But, the general
development above is still applicable, and we still need 
to sample at rate exceeding $b-a$ Hz, and it helps to keep in
mind that in this general case, each sample is actually two real
numbers, not one, since we are sampling a complex-valued signal.