Here is an expansion on Carlo's answer. It is more in the nature of clarifying what is sought as well as a nice parallel.

First another well known permutation of the integers is the the 2-ary array (A054582). For technical reasons is it convenient to index rows and columns starting with $0$

$\left[ \begin {array}{cccccc} 1&2&4&8&16&32\\ 3&6&
12&\mathbf{24}&48&96\\ 5&10&\mathbf{20}&40&80&160\\
7&14&28&56&112&224\\ 9&18&36&72&144&288
\\ 11&22&\mathbf{44}&88&176&352\end {array} \right]$

The analogous question is

Given pairs $(i,j)$ and $(k,\ell)$ find the pair $(p,q)$ giving the position of the sum of the entries in the two given positions.

for example given $(1,3)$ and $(2,2)$ we wish to get back $(5,2)$ since the corresponding entries are $24$ $20$ and $44$

It is not required to find $24,20$ and $44$ nor is it forbidden.

That is one strategy because we easily find

DECODE: The find the entry at position $(i,j),$ multiply $(2i+1)\cdot 2^j$

ENCODE: To find the position of $n$ factor out the twos to get $n=(2i+1)\cdot 2^j$

So then an easy solution to the given problem is decode $(i,j)$, decode $(k,\ell)$, add the results (however one adds) and then encode the sum.

It is relevant to use binary rather than decimal for the array hence

$\left[ \begin {array}{cccccc} 1&10&100&1000&10000&100000
\\ 11&110&1100&11000&110000&1100000
\\ 101&1010&10100&101000&1010000&10100000
\\ 111&1110&11100&111000&1110000&11100000
\\ 1001&10010&100100&1001000&10010000&100100000
\\ 1011&10110&101100&1011000&10110000&101100000 \end{array}\right]$

DECODE: Write $i$ in binary (better yet, use binary) write another $1$ then end with $j$ $0$'s This gives you the entry in binary.

So given $(10,3)$ we write $10$ in binary then a $1$ then $3$ zeros $1010\ 1\ 000$ i.e. $10101000$ Given a second pair we could write a second binary string and then perform binary addition. It is true that $21\cdot 2^3=168=10101000_2$ but in some sense we stayed closer to the given information. Especially if instead of calling it row $10$ we call it row $1010$

ENCODE: Given $n$ convert it to binary (better yet, get it in binary). The number of terminal $0$'s give the column and the head, removing the last $1$ gives the row.

Recall that there are two common ways to get the binary expansion of $i$: go right to left putting a 1 for odd $n$ and replacing it by $\frac{n-1}2$ or $0$ for even $n$ then replacing with $\frac{n}2.$ OR we can go left to right subtracting off the largest possible power of $2$ , say $2^i$ and putting a $1$ then continue considering $2^j$ for $j=i-1,i-2,\cdots$ each time either subtracting and putting a $1$ or just putting a $0$ according as the current value is or is not at least $2^j.$

The second method applied using the Fibonacci numbers $\cdots 13,8,5,3,2,1$ gives the unique Zeckendorf representation of $n$ as a binary vector with no two consecutive $1$'s. (I don't know if there is an easy right to left method.)

Using this representation turns the array of the question

$\left[ \begin {array}{cccccc} 1&2&3&5&8&13\\ 4&7&11
&18&29&47\\ 6&10&16&26&42&68\\ 9&
15&24&39&63&102\\ 12&20&32&52&84&136
\\ 14&23&37&60&97&157\end {array} \right] $

into

$\left[ \begin {array}{cccccc} 1&10&100&1000&10000&100000
\\ 101&1010&10100&101000&1010000&10100000
\\ 1001&10010&100100&1001000&10010000&100100000
\\ 10001&100010&1000100&10001000&100010000&
1000100000\\ 10101&101010&1010100&10101000&101010000
&1010100000\\ 100001&1000010&10000100&100001000&
1000010000&100001000000\end{array} \right]$

If one has the first column then one has the Zeckendorf representation of the entry in position $(i,j)$ and can do the appropriate addition, which is not quite as easy as binary addition but not too bad. And there are formulas such as $\lfloor i \tau^2 \rfloor-1=\lfloor i \tau \rfloor+i-1.$ Here $\tau$ could be replaced by a sufficiently good ratio of Fibonacci numbers. This still leaves some loose ends such as how to figure out the row from the representation.