Structure of the bifluoride anion
Quoting from a comment of your’s:


  Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present.


This sentence is correct for simple two-electron-two-centre bonds; i.e, if each of the two bonds going to hydrogen were to represent a 2e2c bond, hydrogen would have a double negative charge.

However, the $\ce{HF2-}$ anion does not contain two 2e2c bonds. Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$. This can be understood by the following graphic:


Figure 1: Scheme of the 4-electron-3-centre bond in $\ce{HF2-}$. Image taken from Professor Klüfers’ web scriptum of his general and inorganic chemistry course.

The lowest molecular orbital, $\sigma_1$, is a fully bonding $\ce{F-H-F}$ orbital with two electrons. The second, $\sigma_2$ is nonbonding; no (accessable) hydrogen orbital exists that could complete it. The final one, $\sigma_3$, is fully antibonding. This is similar to the orbital scheme of extended linear &#x3c0; systems, where the number of nodal planes increases by 1 with every higher orbital. And it should be understood as such: the nodal plane of $\sigma_2$ traverses the hydrogen atom. Thus, while formally nonbonding, $\sigma_2$ is actually weakly bonding for both $\ce{H-F}$ bonds. If we calculate the overall bond order, we still count the electrons in $\sigma_2$ as nonbonding, giving a bond order of $0.5$ for each $\ce{H-F}$ bond.

The molecule can be understood in 2e2c Lewis terminology by using the following two resonance structures:

$$\ce{F-H\bond{...}F- &lt;-&gt; ^-F\bond{...}H-F}$$

‘Multiple bonds’ going away from a single hydrogen is not restricted to the 4-electron-3-centre bond known in $\ce{HF2-}$; they are also known and well-studied in boranes, although they are typically 2-electron-3-centre bonds in those electron-deficient compounds.