I think you are confused by negative frequencies and what they mean so let me add this explanation.
When you see a spectrum that contains "positive" and "negative" frequencies, each of the frequencies are of the form:
$$e^{j\omega t}$$
Where $\omega$ is the frequency (in this case angular frequency as $2\pi f$ with f being the frequency in Hz.
The general form of $e^{j\phi}$ is a phasor on the complex plane with magnitude one and angle $\phi$, so $e^{j\omega t}$ is simply a phasor with unit magnitude rotating with rate $\omega$. If $\omega$ is positive, the rotation is counter-clockwise: a positive frequency, and if the rotation is negative, the rotation is clockwise and represents a negative frequency. So the concept of positive and negative frequencies only applies to frequencies of the form of the complex exponential and is the direction of rotation.
Consider Euler's identity for a cosine that demonstrates the concept of negative frequencies well:
$$cos(\omega t) = \frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$
Here we get two phasors on the complex plane, one rotating clockwise, one counter-clockwise, and adding the two results in a signal that always stays on the real axis going back and forth sinusoidally.
So therefore a simple cosine wave with a constant frequency would contain one positive (exponential) frequency and one negative (exponential) frequency. We also see from this that ANY real signal must have positive and negative frequencies, AND the negative frequencies MUST be of the opposite phase. Like the cosine example above, each phasor component spinning in a counter-clockwise direction (the positive frequencies) MUST have an equal magnitude phasor spinning in the opposite direction (negative frequency component) in order for the sum to stay on the real axis (which would then be a real signal with no imaginary components. This is said to be "complex conjugate symmetric" when the negative frequency components have equal magnitude and opposite phase compared to the positive frequency components.
In order to IMPLEMENT a signal that does not have a spectrum with this property (positive and negative frequencies not complex conjugate symmetric), you would need to have TWO values in implementation, one to represent the real number and one to represent the imaginary number (or you could do magnitude and phase, either way you need two signals). You see this in the baseband part of your diagram where the real is represented as "I" for "in-phase" and the imaginary is represented as "Q" for quadrature. So in that part of your diagram we are indeed representing a complex signal, and the spectrum at that point can be asymmetric.
I notated your diagram to demonstrate this further. I will call the IQ signal the "baseband signal" and because it is indeed complex, it's spectrum can be assymetric as I have drawn. This complex baseband signal is then digitally upconverted (frequency translated) by multiplying it by $e^{-j\omega t}$ since $e^{-j\omega t} = cos(\omega t)-jsin(\omega t)$ and then taking the real part:
$$(I_1+jQ_1)(I_2-jQ_2)$$
$$= (I_1I_2+Q_1Q_2)+j(I_1Q_2-I_2Q_1)$$
Taking the real part of this results in the form in your block diagram prior to the DAC:
$$(I_1I_2+Q_1Q_2)$$
Given as:
$$I cos(2\pi f_{IF}t) + Q sin(2\pi f_{IF}t)$$
With $I_1$ and $Q_1$ as the I and Q inputs to the digital multiplier and $I_2$ is the cosine waveform into the upper digital multiplier and $Q_2$ is the sine waveform input into the lower digital multiplier in the diagram)
So the signal where we maintain two datapaths (truly a complex signal) will have an asymmetric spectrum while the signal where this is only one datapath must be real and therefore must have a spectrum that is complex conjugate symmetric. I like this diagram as shown as it clarifies that the first signal I circled is indeed complex and has an asymmetric spectrum, while the second one I circled, while it looks like the spectrum is asymmetric if you just considered the positive frequencies alone, is not as we are referring to symmetry between the positive and negative frequency axis.
So we see how negative frequencies exist throughout the entire block diagram, and it is not that the DAC is creating them. The baseband I and Q signal coping out of the encoder is complex and has an assymetric spectrum, this gets digitally upconverted (and is still complex until summed), the output of the summer is now one signal path so must be real and therefore now has a symmetric (complex conjugate) spectrum. The DAC will convert this to the analog domain and the bandpass filter will filter out the images of the DAC (positive and negative--it is a real filter so it too is complex conjugate symmetric in its passband). So assuming the BPF is centered on $f_{IF}$, the analog spectrum would look identical to the spectrum prior to the DAC, extend to $\pm \infty$ while the unique DAC input spectrum extends only to $\pm f_s/2$.
The "image" shown in the diagram are NOT the images due to digital sampling, but the image due to the analog upconversion. This is a real multiplier and as such its output will contain the sum and the difference of the input (review the expansion of $cos(\alpha)cos(\beta)$ to see this), so we get the terms RF+IF and RF-IF at both postiive and negative frequency locations. (However you can also see that we are just frequency translating the same spectrum at the output of the BPF by convolving it with the two frequencies of the RF sinusoid.)
There is more not detailed about the DAC upconversion that I didn't cover but you should be aware of including the option to pass a higher Nyquist zone by setting the bandpass filter on one of the DAC images, and that the DAC itself will introduce a Sinc roll--off in frequency due to the staircase reconstruction.
