Hilbert spaces and unique extensions of linear functions.

Let me simplify things by assuming (as you point out, without loss of generality) that $V$ is closed. Then, indeed, $f(x) = \langle x,y \rangle$ on $V$ for a unique $y \in V$, yielding the immediate extension $h(x) := \langle x,y\rangle$ of $f$ to $H$; one can readily check that $$\|f\|_{V^\ast} = \|y\|_V = \|y\|_H = \|h\|_{H^\ast},$$ so that this extension is norm-preserving.

Now, suppose that $k \in H^\ast$ is another extension of $f$ to $H$, so that $k(x) = \langle x,z\rangle$ for a unique $z \in H$. Then for any $x \in V$, $$0 = f(x) - f(x) = h(x) - k(x) = \langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle,$$ so that $y-z \in V^\perp$. Hence, $y$ and $y-z$ are orthogonal, so that $$\|k\|_{H^\ast} = \|z\|_H = \|y - (y-z)\|_H = \sqrt{\|y\|^2_H + \|y-z\|^2_H}.$$ Thus, $\|k\|_{H^\ast} = \|f\|_{V^\ast}$ if and only if $\sqrt{\|y\|^2_H + \|y-z\|^2_H} = \|y\|_H$, if and only if $\|y-z\| = 0$, if and only if $y =z$, if and only if $h=k$. Hence, $h$ is indeed the unique norm-preserving extension.