If $M$ is a compact Riemannian manifold and $g$ and $\tilde{g}$ are metrics on $M$, then $\frac{1}{C} g \leq \tilde{g} \leq C g$ for $C > 1$

You can prove this in a more direct way. It looks like the proof that in a finite dimensional vector space, all norms are equivalent.

Let $S_gM$ be the unit sphere bundle of $(M,g)$, that is $S_gM = \{ (p,v)\in TM | g_p(v,v)=1 \}$. If $M$ is compact, then $S_gM$ is compact too. The smooth function $f$ on $TM$ defined by $f(p,v)= \tilde{g}_p(v,v)$ is then continuous restricted to $S_gM \subset TM$. Notice $f$ is positive, as every $v\in S_gM$ is non-zero. By compactness, there exist $m,M >0$ such that $m\leqslant f(p,v) \leqslant M$ on $S_gM$. You can chose some constant $C>1$ such that $\frac{1}{C} \leqslant m \leqslant M \leqslant C$, so that on $S_gM$, $\frac{1}{C} \leqslant \tilde{g}_p(v,v)\leqslant C$. By the very definition of $S_gM$, we have that for every $(p,v)\in S_gM$, $$\frac{1}{C}g_p(v,v)\leqslant \tilde{g}_p(v,v) \leqslant Cg_p(v,v)$$ Now, the homogeneity of quadratic forms show that this inequality is true on all $TM$.

The point of this answer is to explain what the question is; the other answer has a perfect proof. As stated in your first quote, $\frac{1}{C} g \leq \tilde{g} \leq C g$ is to be understood in the sense of quadratic forms. This means that for all $x\in M$ and $v\in T_xM$ we have $$ \frac{1}{C} g_x(v,v) \leq \tilde{g}_x(v,v) \leq C g_x(v,v) $$ or equivalently in local coordinates $$ \frac{1}{C} \sum_{i,j}g_{ij}(x)v^iv^j \leq \sum_{i,j}\tilde{g}_{ij}(x)v^iv^j \leq C \sum_{i,j}g_{ij}(x)v^iv^j. $$ This means that the two norms on every $T_xM$ are bi-Lipschitz equivalent and the constant is independent of $x$.

Even when we write $\frac{1}{C} g_{ij} \leq \tilde{g}_{ij} \leq C g_{ij}$, it can be a shorthand for the inequalities in the sense of quadratic forms. That is indeed a much more likely interpretation than a componentwise result.

To underline the importance of working with quadratic forms rather than individual components, let me define three (partial) orders for symmetric square matrices:

• In the sense of quadratic forms: $A\leq_{qf}B$ means that $v^TAv\leq v^TBv$ for all $v$.
• Componentwise: $A\leq_{cw}B$ means that $A_{ij}\leq B_{ij}$ for all indices.
• For all pairs: $A\leq_{p}B$ means that $u^TAv\leq u^TBv$ for all $u$ and $v$.

Now take $$ A = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} $$ and $$ B = \begin{pmatrix} 1&10\\ 10&1 \end{pmatrix}. $$ Clearly $A\leq_{cw}B$, but for $v=(1,-1)$ we have $$ 2 = v^TAv > v^TBv = -18. $$ Thus $A\leq_{cw}B$ does not imply $A\leq_{qf}B$.

In the case of Riemannian metrics proving $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is insufficient, and in general it does not even hold. For example, if $\tilde g$ is the Euclidean metric (the identity matrix) and $g$ is a Riemannian metric with non-zero (perhaps both positive and negative) off-diagonal entries at some point, the componentwise version is false but the version with quadratic forms is still valid.

In general, $A\leq_{p}B$ implies both $A\leq_{qf}B$ (use the same vector twice) and $A\leq_{cw}B$ (choose two basis vectors). While the order given by pairs of vectors implies the correct one, it often fails because because the componentwise one does even though the desired estimate holds true.

What you need is $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$, not $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ or $\frac1Cg\leq_{p}\tilde g\leq_{p}Cg$. Unfortunately your proof that $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$ implies $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is invalid.