Let $(X,\langle\cdot\,,\cdot\rangle_X)$ and $(Y,\langle\cdot\,,\cdot\rangle_Y)$ be nonzero inner product spaces over $\mathbb{C}$.
I wish to know if the following statement is true.
$$\sum_{j=1}^n\sum_{k=1}^n a_{j,k}\langle x_j,x_k \rangle_X=0,\quad\forall\, x_1,\ldots,x_n\in X$$
if and only if
$$\sum_{j=1}^n\sum_{k=1}^n a_{j,k}\langle y_j,y_k \rangle_X=0,\quad\forall\, y_1,\ldots,y_n\in Y$$
where $a_{j,k}\in\mathbb{C},\,\forall\,j,k=1,\ldots,n$.
A particular case of this statement is as follows.
$$\sum_{j=1}^n a_j\Big\|\sum_{k=1}^m b_{j,k}x_k\Big\|_X^2=0,\quad\forall\,x_1,\ldots,x_m\in X$$
if and only if
$$\sum_{j=1}^n a_j\Big\|\sum_{k=1}^m b_{j,k}y_k\Big\|_Y^2=0,\quad\forall\,y_1,\ldots,y_m\in Y$$
where $a_j,b_{j,k}\in\mathbb{C},\,\forall\,j=1\ldots,n;k=1,\ldots,m$.
I don't know if this is true either.
You may need to add some condition on the dimension of the spaces. If both have dimension of at least n, (or identical dimension since then they are isometric anyway) then what you say should work and the proof should look something like this:
Assume that the equation holds for $X$, and let $y_1$ to $y_n$ be in $Y$. Now find $x_1$ to $x_n$ in $X$ such that $$\left\langle x_k,x_l\right\rangle_X = \left\langle y_k,y_l\right\rangle_Y \quad \forall k,l \in \{1,...,n\}$$ (should be doable by some linear algebra, you may need to work out the details. also here the dimension of at least $n$ comes in). Then (I changed the last sum to a sum over $l$) by just the properties of the inner product $$\sum_{j=1}^m\left\langle \sum_{k=1}^n a_{j,k} y_k, \sum_{k=1}^n b_{j,l} y_l \right\rangle_Y = \sum_{j=1}^m \sum_{k=1}^n a_{j,k} \sum_{l=1}^n \overline{b_{j,l}} \left\langle y_k, y_l \right\rangle_Y $$
$$ = \sum_{j=1}^m \sum_{k=1}^n a_{j,k} \sum_{l=1}^n \overline{b_{j,l}} \left\langle x_k, x_l \right\rangle_X = \sum_{j=1}^m\left\langle \sum_{k=1}^n a_{j,k} x_k, \sum_{k=1}^n b_{j,l} x_l \right\rangle_X = 0$$ Which proves your statement.
Concerning a lower amount of dimensions, I think there may be a counterexample, but I did not try to find one yet.