Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base?
You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt $\ce{NH4Cl}$ of which the ammonium cation is acidic, in water we get reaction (1).

$$\ce{NH4+ + H2O &lt;=&gt; NH3 + H3O+}\tag{1}$$

To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant $K_\mathrm a$ as shown in (2). (Wether we use $\ce{H+}$ or $\ce{H3O+}$ is irrelevant.)

$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$

Using the tabulated $\mathrm pK_\mathrm{a}$ value of 9.25, we can calculate the amount of $\ce{NH3}$ in a $\pu{1M}$ solution of $\ce{NH4Cl}$:

$$\begin{align}K_\mathrm a &amp;= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em]
K_\mathrm a (c_0 - [\ce{NH4+}]) &amp;= [\ce{NH4+}]^2\\[0.5em]
[\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &amp;= 0\\
[\ce{NH4+}] &amp;= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\
[\ce{NH4+}] &amp;= \pu{2.37e-5 M}\tag{3}\end{align}$$

Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, $99.998~\%$ are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.

Only if you add both a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.