We will use the infinity or max norm on $V\times V$, that is, for $(u,v)\in V\times V$, $\lVert (u,v)\rVert_{V\times V}=\max\{\lVert u\rVert_V,\lVert v\rVert_V\}$, where the norm on $V$ is the norm induced by the inner product $\langle\cdot,\cdot\rangle$. I can't speak with authority on why this is equivalent to the product topology, but I don't think it should be too difficult to show for a finite product of normed spaces - those are some well-behaved constructs. For a function to be jointly continuous, it means simply that it is continuous in the product topology, rather than only being continuous in each variable separately, with the others fixed. That is, for $f:X\times Y\to Z$, $f$ is jointly continuous if, for all $(x_0,y_0)\in X\times Y$, given any $\epsilon>0$, there exists a $\delta>0$ such that, for all $(x,y)\in X\times Y$, $d_{X\times Y}((x,y),(x_0,y_0))<\delta$ implies $d_Z(f(x,y),f(x_0,y_0))<\epsilon$. So let's do it, using the induced norm as our metric.
Fix an $x_0,y_0\in V$, and let $\epsilon>0$ be given. Choose $\delta=\min\{1,\frac{\epsilon}{2(\lVert y_0\rVert + 1)},\frac{\epsilon}{2(\lVert x_0\rVert+1)}\}$, and note $\lvert\lVert y\rVert_V-\lVert y_0\rVert_V\rvert\leq \lVert y-y_0\rVert_V<1$, by the reverse triangle inequality and our choice of $\delta$. This implies $\lVert y\rVert_V<1+\lVert y_0\rVert_V$.
Let $x,y\in V$, and suppose $\lVert (x,y)-(x_0,y_0)\rVert_{V\times V}=\lVert(x-x_0,y-y_0)\rVert_{V\times V}<\delta$, implying $\lVert x-x_0\rVert_V<\delta$ and $\lVert y-y_0\rVert_V<\delta$. Then
$\lvert\langle x,y\rangle-\langle x_0,y_0\rangle\rvert = \lvert\langle x,y\rangle-\langle x_0,y\rangle+\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$
$\leq \lvert\langle x,y\rangle-\langle x_0,y\rangle\rvert+\lvert\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$
$= \lvert\langle x-x_0,y\rangle\rvert+\lvert\langle x_0,y-y_0\rangle\rvert$
$\leq \lVert x-x_0\rVert_V\lVert y\rVert_V + \lVert x_0\rVert_V \lVert y-y_0\rVert_V$ (by Cauchy-Bunyakovsky-Schwarz inequality)
$< (\frac{\epsilon}{2(\lVert y_0\rVert+1)})(1+\lVert y_0\rVert_V) + (\lVert x_0\rVert_V+1)(\frac{\epsilon}{2(\lVert x_0\rVert+1)})$, (since $\lVert x_0\rVert_V<\lVert x_0\rVert_V+1$, and by our choice of $\delta$)
$= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$.
Best Answer
There is an implicit question about the convergence of the sums, which this answer does not address.
If $\sum_{n=1}^\infty x_n$ converges in the norm of the Hilbert space $H$, then for any $y \in H$, $\sum_{n=1}^\infty \langle x_n, y \rangle$ converges in the base field ($\mathbb{R}$ or $\mathbb{C}$), and $$\langle \sum_{n=1}^\infty x_n , y \rangle = \sum_{n=1}^\infty \langle x_n, y \rangle.$$ As mentioned, this is because the inner product is linear and continuous with respect to the $H$ norm topology (essentially, by Cauchy-Schwarz).
However, if the sum on the right side converges for some $y \in H$, or even for all $y \in H$, it does not follow that $\sum_{n=1}^\infty x_n$ converges in $H$, so the left side may be meaningless. For an example, let $\{e_i\}_{i=1}^\infty$ be an orthonormal set in $H$ (assuming it is infinite dimensional). Let $x_1 = e_1$, and $x_n = e_n - e_{n-1}$ for $n \ge 2$. Then $\sum_{n=1}^\infty x_n = \lim_{i \to \infty} e_i$ does not converge in norm. However, $\sum_{n=1}^\infty \langle x_n, y \rangle = \lim_{i \to \infty} \langle e_i, y \rangle = 0$ for all $y$, by Bessel's inequality.
Effectively, having the right side converge for all $y$ only implies that $\sum_{n=1}^\infty x_n$ converges in the weak topology.