Try to fill in the details in the following outline:
Fact 1: For any chain complex $C_*$ one has that $$\sum_n (-1)^n \operatorname{rank} C_n=\sum_n (-1)^n \operatorname{rank} H_n(C_*).$$
Prove this fact by considering the two short exact sequences $$0\to Z_n \to C_n \stackrel{\partial}{\to} B_{n-1}\to 0$$ and $$0\to B_n\to Z_n \to H_n(C_*)\to 0$$ where $B_n\subset C_n$ are the $n$-boundaries and $Z_n\subset C_n$ the $n$-cycles.
Taking that fact for granted, consider the Mayer-Vietoris sequence corresponding to the triple $(2M; M_1, M_2)$:
$$\cdots \to \underbrace{H_1(2M)}_{C_3}\to \underbrace{H_0(\partial M)}_{C_2}\to \underbrace{H_0(M_1)\oplus H_0(M_2)}_{C_1}\to \underbrace{H_0(2M)}_{C_0}\to 0$$
Since it is a long exact sequence, it is in particular a chain complex with vanishing homology. Define chain modules $C_n$ as indicated above. Fact 1 tells you that $\sum_n (-1)^n \operatorname{rank} C_n=0$.
Write out what that means, rearrange the terms and you will obtain the desired formula $\chi(2M)=2\chi(M)-\chi(\partial M)$.
Fact 2: The $k$-th and $(n-k)$-th Betti numbers of a closed orientable $n$-manifold coincide.
This fact is a corollary of Poincaré Duality. As was hinted in the comment section, use this to prove that $\chi(2M)=0$.
$\newcommand{\Q}{\mathbb{Q}}$Poincaré duality tells you that there are non-degenerate pairings $H^i(M) \otimes H^{n-i}(M, \partial M) \to \Q$ for all $0 \le i \le n$.
Using the long exact sequence of the pair $(M, \partial M)$, the known facts that $H^n(M) = 0$ and $H^n(M,\partial M) = \Q$ (Poincaré duality again), and the description of $H^*(\partial M) = H^*(S^{n-1})$, one can deduce that $H^i(M) \cong H^i(M, \partial M)$ for all $i \le n-1$. The non-degenerate pairing mentioned above thus tells you that $\dim H^i(M) = \dim H^{n-i}(M)$ for all $0 < i < n$.
Write $n = 4k+2$ for convenience.
We can therefore start to compute the Euler characteristic, pairing terms appropriately:
\begin{align}
\chi(M) & = \sum_{i=0}^n (-1)^i \dim H^i(M) \\
& = \dim H^0(M) + \sum_{i=1}^{2k} (-1)^i \dim H^i(M) - \dim H^{2k+1}(M) \\
& \qquad+ \sum_{i=2k+2}^{4k+1} (-1)^i \dim H^i(M) + \dim H^{4k+2}(M) \\
& = 1 + 2 \sum_{i = 1}^{2k} (-1)^i \dim H^i(M) - \dim H^{2k+1}(M).
\end{align}
So we just need to show that $\dim H^{2k+1}(M)$ is even. Because it's the middle dimension, we have a non-degenerate pairing $H^{2k+1}(M) \otimes H^{2k+1}(M, \partial M) \to \Q$. Composing with the isomorphism $H^{2k+1}(M) \cong H^{2k+1}(M,\partial M)$ we get a non-degenerate pairing $H^{2k+1}(M) \otimes H^{2k+1}(M) \to \Q$, given by $\alpha \otimes \beta \mapsto \langle \alpha \cup \beta, [M] \rangle$.
Since $2k+1$ is odd, this pairing is skew-symmetric (and non-degenerate). In particular, $H^{2k+1}(M)$ is a symplectic vector space, and therefore it is necessarily even-dimensional. Plugging this in the formula for $\chi(M)$ above, we finally get that $\chi(M)$ is odd.
Best Answer
Lets suppose that we're working with $M$ a finite dimensional manifold with boundary, and for convenience we'll suppose $\dim M$ is even (If it's not, we would just need to use the fact that $0=-0$ at the end). Let's split $X$ up in the obvious way as the union of two copies of $M$ with a slight thickening in to each other just so their interiors cover $X$ and we satisfy all of the criteria to use Mayer-Vietoris. We'll call them $M_1$ and $M_2$ which intersect on a subspace which is homeomorphic to $\partial M\times [0,1]$. This intersection is homotopy equivalent to the boundary of $M$ so we'll just call it $\partial M$ without confusion.
By Mayer-Vietoris (and the fact that $\partial M$ has dimension strictly less than $n$ so $H_n(\partial M)=0$), we get a long exact sequence $$\dots \to 0 \to H_n(M_1) \oplus H_n(M_2) \to H_n(X) \to H_{n-1}(\partial M) \to \dots$$ $$\dots \to H_1(\partial M) \to H_0(M_1) \oplus H_0(M_2) \to H_0(X) \to H_{0}(\partial M) \to 0$$ (note, I am considering all coefficients of homology groups to be in the rationals).
Now, we need two facts. The first is that the general rank nullity theorem tells us that for a long exact sequence $0\rightarrow V_1\rightarrow\ldots\rightarrow V_r\rightarrow 0$ of finite dimensional vector spaces, we have $$0=\sum_{i=1}^r (-1)^i \dim V_i.$$ The second is that for a topological space $Y$ with definable Euler characteristic, $$\chi (Y)=\sum_{i=0}^\infty (-1)^i\dim (H_i(Y;\mathbb{Q})).$$
Now, because $n$ is even, we get the following sum from the above facts and a little rearranging of the finite sum $$0=[\dim (H_n(M_1))+\dim (H_n(M_1))]-\dim (H_n(X))+\dim (H_{n-1}(\partial M))+\ldots$$ $$\ldots+\dim(H_1(\partial M))-[\dim (H_1(M_1))+\dim (H_1(M_1))]+\dim (H_1(X))$$ $$-\dim (H_0(\partial M))+[\dim(H_0( M_1)) + \dim(H_0(M_2))]-\dim (H_0(X))$$ $$=-\chi (\partial M)+\chi (M)+\chi (M)-\chi (X)$$ $$\Rightarrow \chi (X)=2\chi(M)-\chi(\partial M).$$