Let \(n=\max\{|V(F_i)|:1\leq i\leq m\}\) and, for each \(i\text{,}\) let \(F_i'\) be obtained from \(F_i\) by adding \(n-|V(F_i)|\) isolated vertices. Since the graphs \(F_1,\dots,F_m\) are pairwise non-isomorphic and do not have isolated vertices, the graphs \(F_1',\dots,F_m'\) are pairwise non-isomorphic. For any graph \(G\text{,}\) we have
\begin{equation*}
t(F_i,G)=t(F_i',G)=\frac{\hom(F_i',G)}{|V(G)|^n}.
\end{equation*}
So, for any graphs \(G_1,\dots,G_m\) we have \([t(F_i,G_j)]_{1\leq i,j\leq m}\) can be obtained from \([\hom(F_i',G_j)]_{1\leq i,j\leq m}\) by scaling the \(j\)th column by \(\frac{1}{|V(G_j)|^n}\) and so if one of these matrices is non-singular, then so is the other. Thus, it suffices to work with the latter matrix.
For \(1\leq i\leq m\) and positive integers \(x_1^i,\dots,x_n^i\text{,}\) we let \(G_i(x_1^i,\dots,x_n^i)\) be a graph obtained from \(F_i'\) by blowing up the \(k\)th vertex into a set of size \(x_k^i\) for all \(1\leq k\leq n\text{.}\) Then \(\det\left([\hom(F_i',G_j(x_1^j,\dots,x_n^j))]_{1\leq i,j\leq m}\right)\) is a polynomial in \(nm\) variables. If we can show that this polynomial is not equal to the zero polynomial, then there must exist a choice for the variables (in fact, infinitely many) such that it is non-zero, which will complete the proof.
So, it all boils down to showing that the determinant is not equal to the zero polynomial. Let \(\hom(F_i',G_j(x_1^j,\dots,x_n^j)) = p_{i,j}(x_1^j,\dots,x_n^j) + q_{i,j}(x_1^j,\dots,x_n^j)\) where \(p_{i,j}(x_1^j,\dots,x_n^j)\) counts the homomorphisms in which every pair of vertices of \(F_i'\) are mapped to vertices of \(G_j(x_1^j,\dots,x_n^j)\) which correspond to different vertices of \(F_j'\text{;}\) i.e. the composition of the homomophism and the natural projection of \(G_j(x_1^j,\dots,x_n^j)\) onto \(F_j'\) is injective. If we say that a monomial is “multilinear” if every variable within it has exponent 1 (e.g. \(x_1x_2\) is multilinear and \(x_1^2x_2\) is not), then \(p_{i,j}(x_1^j,\dots,x_n^j)\) consists of the multilinear term of the polynomial \(\hom(F_i',G_j(x_1^j,\dots,x_n^j))\text{.}\) Also, the multilinear term of the determinant of \([p_{i,j}(x_1^j,\dots,x_n^j) + q_{i,j}(x_1^j,\dots,x_n^j)]_{1\leq i,j\leq k}\) is precisely equal to the determinant of \([p_{i,j}(x_1^j,\dots,x_n^j)]_{1\leq i,j\leq k}\text{.}\) So, if we can show that this determinant is non-zero, then we will know that our polynomial is not equal to the zero polynomial (since it contains a multilinear term with non-zero coefficient). Without loss of generality, we can assume that \(F_1',\dots,F_k'\) are listed so that \(|E(F_1')|\leq |E(F_2')|\leq \cdots\leq |E(F_k')|\text{.}\) Since the graphs \(F_1',\dots,F_k'\) are pairwise non-isomorphic, this implies that \(p_{i,j}(x_1^j,\dots,x_n^j)=0\) whenever \(i>j\text{.}\) Also, \(p_{i,i}(x_1^i,\dots,x_n^i)>0\) for any \(i\) by considering the identity map. So, \([p_{i,j}(x_1^j,\dots,x_n^j)]_{1\leq i,j\leq k}\) is a lower triangular matrix with no zeros on the diagonal and therefore it has non-zero determinant. This completes the proof.