$\newcommand{\om}{\omega}$Let me answer your specific question.
The proof is similar to that of the uniform boundedness principle for linear functionals, but here using the identity \begin{equation} \begin{aligned} 4T_m(s,t)&=T_m(x+s,y+t)-T_m(x+s,y-t) \\ &-T_m(x-s,y+t)+T_m(x-s,y-t) \end{aligned} \tag{10}\label{10} \end{equation} for all $s,t,x,y$ in $E$.
Indeed, for natural $n$ let \begin{equation} F_n:=\{(v,w)\in E\times E\colon\,\sup_m|T_m(v,w)|\le n\}. \end{equation} Because the $T_m$'s are continuous, the sets $F_n$ are closed. Also, the condition \begin{equation} \lim_m T_m(v,w)=T(v,w) \tag{20}\label{20} \end{equation} for all $v,w$ in $E$ implies that $\bigcup_n F_n=E$. So, by the Baire category theorem, for some natural $n$, some $(x,y)\in E\times E$, and some balanced neighborhood $U$ of $0$ in $E$ we have \begin{equation} F_n\supseteq (x+U)\times (y+U). \end{equation} So, by \eqref{10}, $|T_m(s,t)|\le n$ for all $m$ and all $(s,t)\in U\times U$, and hence, in view of \eqref{20}, $|T(s,t)|\le n$ for all $(s,t)\in U\times U$.
Thus, $T$ is bounded on a neighborhood of $(0,0)$ and hence continuous. $\quad\Box$
The same kind of argument holds for $k$-linear forms for any natural $k$. Then identity \eqref{10} will have to be replaced by the identity \begin{equation} 2^k\, T_m(s_1,\dots,s_k) \\ =\sum_{(\om_1,\dots,\om_k)\in\{-1,1\}^k} (-1)^{\om_1+\dots+\om_k}T_m(x_1+\om_1 s_1,\dots,x_k+\om_k s_k) \tag{10a}\label{10a} \end{equation} for all $s_1,\dots,s_k,x_1,\dots,x_k$ in $E$. In turn, identity \eqref{10a} can be checked by writing the $k$-fold sum as a $k$-fold iterated sum.