For finitely generated projective modules $X$, the evaluation map $X\to X^{tt}$ (defined by $x\mapsto[\varphi\mapsto\varphi(m)]$) is a natural isomorphism.
So $P_1\to P_0$ is isomorphic to $P_1^{tt}\to P_0^{tt}$, which in turn is isomorphic to the direct sum of $E_1'^t\xrightarrow{u^t}E_0'^t$ and $E_1''^t\xrightarrow{v^t}E_0''^t$.
But $v$ is an isomorphism, so $v^t$ is an isomorphism and so its cokernel is zero. So $M$, which is the cokernel of $P_1\to P_0$, is also the cokernel of the first summand $E_1'^t\xrightarrow{u^t}E_0'^t$.
(1) It is true in general that $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n))=0$.
To see this, consider the beginning of the Koszul complex for $(R,x_1,\dots,x_n)$:
$$
\bigwedge^2R^{\oplus n}\to R^{\oplus n}\xrightarrow{
\begin{pmatrix}
x_1&\dots&x_n
\end{pmatrix}
}R\to R/(x_1,\dots,x_n)\to 0.
$$
This is in general not exact at $R^{\oplus n}$, but if we add some free $R$-module $F$ to the leftmost term, we get an $R$-free resolution of $R/(x_1,\dots,x_n)$:
$$
F \oplus \bigwedge^2R^{\oplus n}\to R^{\oplus n}\to R\to R/(x_1,\dots,x_n)\to 0.
$$
Cutting out $R/(x_1,\dots,x_n)$ and tensoring with $M$, we obtain a complex
$$
(M\otimes_R F)\oplus \biggl(M\otimes_R\bigwedge^2R^{\oplus n}\biggr)\to M^{\oplus n}\to M
$$
whose cohomology at the middle gives $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n)$.
Now, since $x_1,\dots,x_n$ is an $M$-regular sequence, the Koszul complex for $(M,x_1,\dots,x_n)$ is exact.
Therefore the above complex is exact at the middle, so we get $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n))=0$.
(2) There is an example such that $\operatorname{Tor}_2^R(M,R/(x_1,x_2))\neq 0$.
Namely, let $k$ be a field and set $R=k[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, $M=R/(X)\simeq k[Y,Z]_{(Y,Z)}$.
Then $Y,Z$ is obviously an $M$-regular sequence, but it is not an $R$-regular sequence.
One can check that the following sequence gives an $R$-free resolution of $M$:
$$
R^{\oplus 3}
\xrightarrow{
\begin{pmatrix}
Z&X&0\\
-Y&0&X
\end{pmatrix}
}
R^{\oplus 2}
\xrightarrow{
\begin{pmatrix}
Y&Z
\end{pmatrix}
}
R\xrightarrow{X}R\to M\to 0.
$$
Cutting out $M$ and tensoring with $R/(Y,Z)$, we obtain a complex
$$
R/(Y,Z)^{\oplus 3}\xrightarrow{
\begin{pmatrix}
0&X&0\\
0&0&X
\end{pmatrix}
}
R/(Y,Z)^{\oplus 2}\xrightarrow{
\begin{pmatrix}
0&0
\end{pmatrix}
}
R/(Y,Z)\xrightarrow{X}R/(Y,Z)
$$
whose cohomology at $R/(Y,Z)^{\oplus 2}$ gives $\operatorname{Tor}_2^R(M,R/(Y,Z))$.
Therefore we get $\operatorname{Tor}_2^R(M,R/(Y,Z))\simeq k^{\oplus 2}\neq 0$.
Best Answer
Proposition 12.13 of Leuschke and Weigand's Cohen Macaulay Representations has a proof of this for general commutative rings.