Partial answer for linear models: 

In linear models, when using the chi-square version of the Wald test, it is *nearly* the case for large $n$ when the null is true (and, afaik, under local alternatives). For the F-test version it is not true but requires (which is precisely what the chi-square version does) multiplication with a correction for the number of hypotheses tested.

Remaining differences stem from different ways of estimating the error variances as e.g. discussed in the links. (So to return to your purpose of validating code, "close" is maybe too vague to be helpful.) More specifically, the Wald statistics in the linear model for $M1$ or $M0$ vs. $M2$ use the error variance estimate for the large model $M2$, whereas the statistic for $M0$ vs. $M1$ uses the error variance estimate based on model $M1$.

In formula, from https://stats.stackexchange.com/questions/449494/ranking-of-wald-lr-and-score-statistic-in-the-normal-linear-regression-model, we have (with $X_j$ the regressor matrix of model $j$) that we can write the Wald statistic in the linear model for $M1$ vs. $M2$ as
$$
\begin{eqnarray}
\mathcal{W}_{12}&=&n\frac{y'(I-P_{X_1})y-y'(I-P_{X_2})y}{y'(I-P_{X_2})y}\label{waldproj}
\end{eqnarray}
$$
Correspondingly, the Wald statistic for $M0$ vs. $M1$ is (and analogously for $M0$ vs. $M2$)
$$
\begin{eqnarray}
\mathcal{W}_{01}&=&n\frac{y'(I-P_{X_0})y-y'(I-P_{X_1})y}{y'(I-P_{X_1})y}\label{waldproj2}
\end{eqnarray}
$$
such that, even though addiding the numerator terms of $\mathcal{W}_{01}$ and $\mathcal{W}_{12}$ would indeed allow us to cancel out $y'(I-P_{X_1})y$, we cannot simply do so as the test statistics do not share the same denominator. 

Hence, we do *not* have $\mathcal{W}_{01}+\mathcal{W}_{12}=\mathcal{W}_{02}$. 

Remark 1: if we knew the error variance, the result would be true, as Wald and LR then are identical, cf. https://stats.stackexchange.com/questions/276192/exact-equivalence-of-lr-and-wald-in-linear-regression-under-known-error-variance.)

Remark 2: If the null is not true, the sum need not even be close to $\mathcal{W}_{02}$: from https://stats.stackexchange.com/questions/449494/ranking-of-wald-lr-and-score-statistic-in-the-normal-linear-regression-model consider $M1$ and $M2$, letting
$$x_{12}:=\frac{y'(I-P_{X_1})y/n}{y'(I-P_{X_2})y/n}=\hat\sigma^2_{M1}/\hat\sigma^2_{M2}.$$

Similarly, 
$$x_{01}:=\frac{y'(I-P_{X_0})y/n}{y'(I-P_{X_1})y/n}=\hat\sigma^2_{M0}/\hat\sigma^2_{M1}.$$

Then, $n(x_{12}-1)=\mathcal{W}_{12}$ and $n(x_{01}-1)=\mathcal{W}_{01}$. If the null $M0$ is true, $x_{ij}\to1$ as all models consistently estimate the error variance $\sigma^2$. 

One might then go on to prove that the sums of the $\chi^2$ random variables of the two submodels is indeed asymptotically equivalent to the random variable associated with $M02$ (not done here).

If, however, $M2$ is true, $\hat\sigma^2_{M0}$ and $\hat\sigma^2_{M1}$ will not be consistent estimators of $\sigma^2$ anymore. Denote $\sigma_0^2=\text{plim}\hat\sigma^2_{M0}$ and $\sigma_1^2=\text{plim}\hat\sigma^2_{M1}$.

Hence, 
$$x_{01}-1\to_p \frac{\sigma_0^2}{\sigma_1^2}-1$$ and $$x_{12}-1\to_p \frac{\sigma_1^2}{\sigma^2}-1,$$ so that 
$$\mathcal{W}_{01}+\mathcal{W}_{12}\sim n\left(\frac{\sigma_0^2}{\sigma_1^2}+\frac{\sigma_1^2}{\sigma^2}-2\right)$$
while
$$
\mathcal{W}_{02}\sim n \left(\frac{\sigma_0^2}{\sigma^2}-1\right)
$$ 

Remark 3: That the result is true under the null (s.th. $\mathcal{W}_{ij}=O_p(1)$) when $n$ is large can also be motivated via asymptotic equivalence of $\mathcal{W}$ and $LR$: from $\mathcal{W}_{ij}=n(x_{ij}-1)$ and $LR_{ij}=n\log(x_{ij})$, write $LR_{ij}=n\log(1+\mathcal{W}_{ij}/n)$. A Taylor expansion around 1 yields 
$$
\begin{eqnarray*}
LR_{ij}&=&n[\mathcal{W}_{ij}/n+o_p(\mathcal{W}_{ij}/n)]\\
&=&n[\mathcal{W}_{ij}/n+o_p(O_p(1)/n)]\\
&=&n[\mathcal{W}_{ij}/n+o_p(n^{-1})]\\
&=&\mathcal{W}_{ij}+o_p(1)
\end{eqnarray*}
$$

Remark 4: For those, like me, needed a second to see why the result is "obviously" correct for LR: note that we can rewrite
$$
LR_{02}=n\log(x_{02})=n\log(x_{01}x_{12})=n\log(x_{01})+n\log(x_{12})=LR_{01}+LR_{12}
$$

Remark 5: Since $\mathcal{W}_{ij}=O_p(1)$ also under local alternatives, Remark 3 should also go through under such local alternatives.

Here is an illustration where the null *is* true:

```r
library(lmtest)
library(sandwich)

n <- 3000
X1 <- rnorm(n)
X2 <- rnorm(n)
y <- rnorm(n)

M0 <- lm(y~1)
M1 <- lm(y~X1)
M2 <- lm(y~X1+X2)

Wald.M1M0 <- waldtest(M1,M0)$F[2]
Wald.M2M1 <- waldtest(M2,M1)$F[2]
Wald.M2M0 <- waldtest(M2,M0)$F[2]

Wald.M2M0
Wald.M1M0 + Wald.M2M1 # only close if we multiply the previous line by 2

LR.M1M0 <- lrtest(M1,M0)$Chisq[2]
LR.M2M1 <- lrtest(M2,M1)$Chisq[2]
LR.M2M0 <- lrtest(M2,M0)$Chisq[2]

LR.M2M0
LR.M1M0 + LR.M2M1 # the same

Wald.M1M0 <- waldtest(M1,M0, test="Chisq")$Chisq[2]
Wald.M2M1 <- waldtest(M2,M1, test="Chisq")$Chisq[2]
Wald.M2M0 <- waldtest(M2,M0, test="Chisq")$Chisq[2]

Wald.M2M0
Wald.M1M0 + Wald.M2M1 # close