Suppose $\varphi:G\to GL(V)$ is a representation and that $V = V_1\oplus V_2$, where $V_1$ and $V_2$ are $G$-invariant subspaces. Let $\varphi^{(i)}$ be the restriction of $\varphi$ to $V_i$. With the direct sum of two representations $\varphi^{(1)}\oplus \varphi^{(2)} : G \to GL(V_1\oplus V_2)$ defined as
$$ (\varphi^{(1)}\oplus \varphi^{(2)})_g (v_1,v_2) = (\varphi^{(1)}_g(v_1),\varphi^{(2)}_g(v_2)), $$
for all $g\in G$, how do you show that $\varphi$ is equivalent to $\varphi^{(1)}\oplus\varphi^{(2)}$?
I am able to show that if $B_i$ is a basis for $V_i$, so that $B=B_1\cup B_2$, then
$$ [\varphi_g]_B = \left( \begin{matrix} [\varphi_g^{(1)}]_{B_1} & 0\\ 0 & [\varphi_g^{(2)}]_{B_2} \end{matrix} \right). $$
But order to show that they are equivalent I need to show that there is an invertible linear map $T:V\to V$ such that $\varphi = T(\varphi^{(1)}\oplus\varphi^{(2)})_gT^{-1}$, for all $g\in G$, and I am not sure how to proceed.
I find it very confusing that the direct product of two representations is a pair of linear maps (and also that it takes two arguments), because I am not sure what $T(\varphi^{(1)}\oplus\varphi^{(2)})_gT^{-1}$ means in this case.
Consider the map$$\begin{array}{rccc}T\colon&V_1\oplus V_2&\longrightarrow&V\\&(v_1,v_2)&\mapsto&v_1+v_2.\end{array}$$It is an isomorphism. Besides, for each $g\in G$,\begin{align}T\bigl((\varphi^{(1)}\oplus\varphi^{(2)})_g(v_1,v_2)\bigr)&=T\bigl(\varphi^{(1)}_g(v_1),\varphi^{(2)}_g(v_2)\bigr)\\&=\varphi^{(1)}_g(v_1)+\varphi^{(2)}_g(v_2)\\&=\varphi_g(v_1+v_2)\\&=\varphi_g\bigl(T(v_1,v_2)\bigr).\end{align}