A question about net ionic equations
To answer your question it is important to know which compound dissociate completely in water when in aqueous solution. For example, consider following reaction:

$$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -&gt; AgCl_{(s)} + NaNO3_{(aq)}} \tag1$$

When ionic compound dissolves in water, each molecule separates to corresponding ions, which influenced by water. All of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ are ionic compounds, which are soluble in water. $\ce{AgCl}$ is, onthe other hand, an ionic compound, but it is insoluble in water (refer to solubility rules).  Hence the reaction $(1)$ can be rewrite as follows:

$$\ce{Ag+_{(aq)} + NO3-_{(aq)} + Na+_{(aq)} + Cl-_{(aq)} -&gt; AgCl_{(s)} + Na+_{(aq)} + NO3-_{(aq)}} \tag2$$

Like in algebra, you can canceled out common ions from each side of the equation $(2)$, which leaves:

$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -&gt; AgCl_{(s)} } \tag3$$

We called the equation $(3)$ a net ionic equation. 

Now, let's look at your equation:

$$\ce{HCl_{(aq)} + NaF_{(aq)} -&gt; HF_{(aq)} + NaCl_{(aq)}} \tag4$$

Now, $\ce{NaF}$ and $\ce{NaCl}$ are ionic compounds, which are soluble in water, hence they separated into ions in aqueous medium. How about $\ce{HCl}$, and $\ce{HF}$? They are soluble in water, but are they ionic? Since both are acids, to answer this question, we should look into their respective $\mathrm{p}K_\mathrm{a}$ values (or $K_\mathrm{a}$ values). Ref.1 is listed followings: 

$$
\begin{array}{c|ccc}
\text{acid/base (chemical formula)} &amp; \mathrm{p}K_\mathrm{a} &amp; K_\mathrm{a} &amp; \\
\hline
\text{hydrogen iodide }(\ce{HI}) &amp; -10.0 &amp; 1.0 \times 10^{10} \\
\text{hydrogen bromide }(\ce{HBr}) &amp; -9.0 &amp; 1.0 \times 10^{9} \\
\text{hydrogen chloride }(\ce{HCl}) &amp; -8.0 &amp; 1.0 \times 10^{8} \\
\text{hydronium ion }(\ce{H3O+}) &amp; -1.7 &amp; 50.1 \\
\text{hydrogen fluoride }(\ce{HF}) &amp; 3.2 &amp; 6.3 \times 10^{-4} \\
\text{acetic acid }(\ce{CH3CO2H}) &amp; 4.7 &amp; 2.0 \times 10^{-5} \\
\hline
\end{array}
$$

Accordingly, $\ce{HCl}$ should be ionic in aqueous solutions and should dissolve in water according to following equation:

$$\ce{HCl_{(aq)} + H2O_{(l)} -&gt; H3O+_{(aq)} + Cl-_{(aq)}} \tag5$$

I put forward arrow because it'd dissociate completely, since $K_\mathrm{a} = \frac{\ce{[H3O+][Cl-]}}{\ce{[HCl]}} = 1.0 \times 10^{8}$.

However, $\ce{HF}$ is a different story. Since its  $K_\mathrm{a} = 6.3 \times 10^{-4} \lt \lt 1$, its dissociation is not complete, and most of $\ce{HF_{(aq)}}$ molecules stay intact (keep its covalent nature):

$$\ce{HF_{(aq)} + H2O_{(l)} &lt;=&gt; H3O+_{(aq)} + F-_{(aq)}} \tag6$$

Note that $K_\mathrm{a}^\ce{HF} = \frac{\ce{[H3O+][F-]}}{\ce{[HF]}} = 6.3 \times 10^{-4}$, and hence $\ce{[HF]} \approx 1.6 \times 10^{3}\ce{[H3O+][F-]}$. Like sugar molecules, which are soluble in water but did not dissociate, $\ce{HF}$ stays as $\ce{HF_{(aq)}}$ in water. Thus, the equation $(4)$ can be rewritten as:

$$\ce{H+_{(aq)} + Cl-_{(aq)} + Na+_{(aq)} + F-_{(aq)} -&gt; HF_{(aq)} + Na+_{(aq)} + Cl-_{(aq)}} \tag7$$

When you made common ions cancelled out, the net ionic equation renains as:

$$\ce{H+_{(aq)} + F-_{(aq)} -&gt; HF_{(aq)} }$$

References:


K. Peter C. Vollhardt, Neil E. Schore, In Organic Chemistry: Structure and Function; Fifth Edition; W. H. Freeman &amp; Co.: New York, NY, 2007 (ISBN: 0-7167-9949-9).