The square of the Frobenius norm of an $n\times d$ matrix $X$ is equal to Tr($X^TX$).
$$||X||_F^2 = \mathrm{Tr}(X^TX).$$
Then in the case of matrix involving addition or subtraction would be, for example,
$$||X-Y||_2^{2} = \mathrm{Tr}((X-Y)^T(X-Y)).$$
How do we solve in this case? Can we just do as follows?
\begin{align} ||X-Y||_2^{2} &= \mathrm{Tr}((X-Y)^T(X-Y)) \\ &= \mathrm{Tr}((Y^T-X^T)(X-Y))\\ &=\mathrm{Tr}(Y^TX-Y^TY-X^TX+X^TY). \end{align}
I'm not sure if this is mathematically correct.
The trace is linear, hence we have
\begin{align} \operatorname{Tr}\big ((X - Y)^\intercal (X - Y)\big) &= \operatorname{Tr}\big ((X^\intercal - Y^\intercal)(X - Y)\big ) \\ &= \operatorname{Tr} (X^\intercal X - X^\intercal Y - Y^\intercal X + Y^\intercal Y )\\ &=\operatorname{Tr}(X^\intercal X) - \operatorname{Tr}(X^\intercal Y)- \operatorname{Tr}(Y^\intercal X) + \operatorname{Tr}(Y^\intercal Y). \end{align}
In general, you can't simplify this result any further.