Image sharing without data overhead
A Flaw

If you try working through your example concretely then you should see at least one issue with decoding.  Consider values mod $8$:

$
A = { 3 } \\
B = {7}   \\
secret = {5 }
$

Using your equations we have:

$
Q = 3 + 7 + 5 = 7 \\
U = 3 - 7 + 5 = 1 
$

And for decoding:

$
B = (U - Q) / 2 \\
B = (1 - 7) / 2 \\
B = -6 / 2 ({\text mod} 8)
$

But what is $\frac{-6}{2}$ modulo 8?  How do we get a unique answer here?

Alternatives

Your constraints are a little unclear, but if you're OK with storing $(1 + {\text nrImages})*|{\text image}|$ (A, B, and secret) then use linear integer secret sharing in a traditional manner (below). Obviously this is a little silly seeing as you could use any symmetric encryption algorithm, given that you have a secret.

If you only want to store ${\text nrImages}|image|$ data, and not the secret, then that is harder as you'll have to derive the secret from the combined shares.  For example we could have a small feistel network:

$
Q0 = A + PRF(B)  \\
U  = B + PRF(Q0) \\
Q  = Q0 + PRF(U)
$

Now $U$ and $Q$ are your secrets, protected ONLY by the entropy of A and B and the strength of your PRF.  The entropy in the images could actually be increased without visually altering the typical image.

To decrypt:

$
Q0 = Q - PRF(U)  \\
B  = U - PRF(Q0) \\
A = Q0 - PRF(B)
$

Basic application of LISS

For LISS just generate a random secret, $s$, and compute:

$
Q = A + B + S \\
U = B + S
$

Then, once the shares and the secrets are collected, you can reverse the process via:

$
A = Q - U \\
B = U - S
$