Is there a feasible way to reconstruct the key from the (Triple-)DES round key?
DES key schedule takes the 64-bit key and PC-1 (Permuted Choice 1) discards the parity check bits and applies a permutation to the remaining.

Then for the rest they key bits are divided into 28-bit halves and for the each round


They are rotated left one or two specified for each round. Then
48-bit bits are selected by PC-2 (Permuted Choice 2) which is another permutation that discards some bits.



\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Number of Round} &amp; 1  &amp; 2 &amp; 3&amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10 &amp; 11 &amp; 12 &amp; 13 &amp; 14 &amp; 15 &amp; 16\\ \hline
\text{Number of Left rotations} &amp; 1 &amp; 1 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 1 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 1 \\ \hline
\end{array}



  Is there a known way to use these round keys to reconstruct the original secret key from it?


Giving a round key of DES, apply reverse PC-2. With this, you will get the 48-bit of the 56 bits key. You can either complete this by using brute-force for the remaining key bits or use another round key to fill the missing 8 key bits. 


  Is there a feasible way to reconstruct the key from the (Triple-)DES round key?


For the Triple DES, you need round keys for each different DES key if the keys are independently generated.

Though the reversibility of a key schedule is not really required, for older systems non-reversible key schedule makes the decryption require more process time or more memory.