Converting karnaugh maps to Boolean expressions
What you understood is correct. Grouping mintems results in sum of product (SOP) form or the AND-OR form as shown in (1) 
$$Y = a_1a_2a_3 + b_1b_2b_3 + \cdots + \omega_1\omega_2\omega_3\tag1$$
Calculating \$\overline{Y}\$ using De-Morgan's theorem,
$$\overline{Y} = (\overline{a_1a_2a_3} )\  (\overline{b_1b_2b_3})  \cdots  (\overline{\omega_1\omega_2\omega_3})$$
From this Y can be written as
$$Y = \overline{(\overline{a_1a_2a_3} )\  (\overline{b_1b_2b_3})  \cdots  (\overline{\omega_1\omega_2\omega_3})}\tag2$$

Which is the NAND-NAND form since variables in a group are NANDed together and NANDed again to obtain Y. So the AND-OR form and NAND-NAND form are equivalent. 

Point to be noted (from(1) and (2) )is that a AND-OR circuit can be converted to NAND-NAND circuit just by replacing the AND and OR gates with NAND gates without changing any interconnections.

So grouping "1"s from K-map allow us to form AND-OR and NAND-NAND easily.

Similarly, grouping maxterms produces POS or OR-AND form:
$${Y} = ({a_1+a_2+a_3} )\  ({b_1+b_2+b_3})  \cdots  ({\omega_1+\omega_2+\omega_3})\tag3$$

Using De-Morgan's theorem it can be proved that OR-AND is equivalent to NOR-NOR form.

So grouping "0"s from K-map allow us to form OR-AND and NOR-NOR easily.

UPDATE:  

From K-map you have to find SOP or POS then you can directly implement using NAND-NAND or NOR-NOR circuit.  

For example, assume that from K-map you obtained the following SOP from K-Map
$$Y = abc + ab\overline{c} + \overline{a}bc$$

You can implement this using AND-OR logic as:



Same way it can be implemented using NAND-NAND as,



So If you can directly draw AND-OR circuit from K-map (SOP) then you can draw NAND-NAND just by replacing the gates with NAND. Similarly, given POS, NOR-NOR circuit can be drawn directly.