Let $U$, $V$, $W$ be real finite-dimensional vector spaces, all equipped with inner products. If $b : U \times V \to W$ is a bilinear map, we can define its operator norm as $$ |b| = \sup_{|u|=1,|v|=1} |b(u,v)|. $$ The map $b$ induces a unique linear map $\hat b : U \times V \to W$ via $\hat b(u \otimes v) = b(u, v)$. The tensor product also inherits an inner product from the ones on $U$ and $V$. We can thus consider the operator norm $$ |\hat b| = \sup_{|x|=1} |\hat b(x)|. $$ Since $b(u,v) = \hat b(u \otimes v)$ and $|u \otimes v| = |u| \, |v|$, we have $|b| \leq |\hat b|$. If $(u_1, \ldots, u_n)$ and $(v_1, \ldots, v_m)$ are orthonormal bases of $U$ and $V$, and $x \in U \otimes V$, we can write $x = \sum_{j,k} a_{jk} u_j \otimes v_k$ for some real numbers $a_{jk}$. If $|x| = 1$, then $\sum_{j,k} |a_{jk}|^2 = 1$. Then $$ |\hat b(x)|^2 \leq \sum_{j,k} |a_{jk}|^2 |b(u_j, v_k)|^2 \leq |b|^2, $$ so we conclude that $|b| = |\hat b|$.
This is all very good. Now let's apply it to the bilinear form $\mathbb R^2 \times \mathbb R^2 \to \mathbb R$ defined by the Euclidean inner product. Its operator norm is $1$ by Cauchy-Schwarz. In the orthonormal basis $(e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2)$ of $\mathbb R^2 \otimes \mathbb R^2$, the matrix of the induced linear map is $$ B = \begin{pmatrix} 1 & 0& 0 & 1 \end{pmatrix}. $$ The operator norm of this is equal to the square root of the largest eigenvalue of $$ B^tB = \begin{pmatrix} 1 & 0& 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0& 0 & 1 \end{pmatrix}. $$ That eigenvalue is $2$, corresponding to the only nonzero eigenvector $(1, 0,0, 1)$, and $\sqrt 2 \not= 1$.
What is wrong with the argument in the first paragraph?