Recall that the t-statistic on the $k$-th coefficient (for testing $\beta_k=0$ assuming homoskedasticity) in a linear regression is given by $$ t=\frac{b_k}{\sqrt{s^2(X'X)^{-1}_{kk}}}, $$ where $(X'X)^{-1}_{kk}$ denotes the $k$-th diagonal element of $(X'X)^{-1}$, so that the squared t-ratio is $$ t=\frac{b_k^2}{s^2(X'X)^{-1}_{kk}}. $$ This is the same as your $$ \frac{b_k^2(x_k'M_1x_k)}{s^2}=\frac{b_k^2}{s^2/(x_k'M_1x_k)} $$ due to the Frisch-Waugh-Lovell theorem discussed in for example Chapter 3 of Greene's textbook.

Recall that the t-statistic on the $k$-th coefficient (for testing $\beta_k=0$ assuming homoskedasticity) in a linear regression is given by $$ t=\frac{b_k}{\sqrt{s^2(X'X)^{-1}_{kk}}}, $$ where $(X'X)^{-1}_{kk}$ denotes the $k$-th diagonal element of $(X'X)^{-1}$, so that the squared t-ratio is $$ t=\frac{b_k^2}{s^2(X'X)^{-1}_{kk}}. $$ This is the same as your $$ \frac{b_k^2(x_k'M_1x_k)}{s^2}=\frac{b_k^2}{s^2/(x_k'M_1x_k)} $$ due to the Frisch-Waugh-Lovell theorem discussed in for example Chapter 3 of Greene's textbook, or more specifically/directly, as an application of the partitioned inverse lemma:

For $A\;(m \times m)$, $B\;(m \times n)$, $C\;(n \times m)$ and $D\;(n \times n)$ we have that, provided the respective inverses exist, $$ \left(% \begin{array}{cc} A & B \\ C & D \\ \end{array}% \right)^{-1}= \left(% \begin{array}{cc} A^{-1} + A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \\ \end{array}% \right). $$ For $A=X_1'X_1$ (with $X_1$ the other regressors except $x_k$), $B=X_1'x_k$, $C=x_k'X_1$ and $D=x_k'x_k$, we can plug in for the lower-right element of the inverse to directly obtain that $$ (D-CA^{-1}B)^{-1}=(X'X)^{-1}_{kk}=1/(x_k'M_1x_k). $$