Given $|\overrightarrow{a}|=8$ and $|\overrightarrow{b}|=12$, and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $45^{\circ}$, we can calculate the dot product $\overrightarrow{a} \cdot \overrightarrow{b}$ using the formula for the dot product in terms of magnitudes and the cosine of the angle between the vectors. The cosine of $45^{\circ}$ is $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$. Therefore, we have:

$$\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| \cdot |\overrightarrow{b}| \cdot \cos(45^{\circ}) = 8 \times 12 \times \frac{\sqrt{2}}{2} = 96 \times \frac{\sqrt{2}}{2} = 48\sqrt{2}$$

The projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is given by the formula:

$$\text{proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{{\overrightarrow{a} \cdot \overrightarrow{b}}}{{|\overrightarrow{b}|^2}} \cdot \overrightarrow{b}$$

Substituting the values we have:

$$\text{proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{{48\sqrt{2}}}{{12^2}} \cdot \overrightarrow{b} = \frac{{48\sqrt{2}}}{{144}} \cdot \overrightarrow{b} = \frac{{\sqrt{2}}}{3}\overrightarrow{b}$$

Therefore, the projection vector of vector $\overrightarrow{a}$ onto $\overrightarrow{b}$ is $\boxed{\frac{\sqrt{2}}{3}\overrightarrow{b}}$.