In Theorem 6.32. pg 355 of Rotman's Hom. Alg., he proves that two different construction of Torsions conincide,
$$Tor_n^R(A,B) \cong tor_n^R(A,B)$$
where
If $B$ is a left $R$-module and $T = - \otimes_R B$, define left derived functor $$Tor_n^R(-, B) = L_nT.$$
If $A$ is a right $R$-module and $T=A \otimes_R - $, define $$tor_n^R(A,-) = L_nT.$$
It is a diagrammatic proof so I hope you may have a look at link. There is one part of proof which I cannot understand:
$W=Tor_1(K_{i-1},V_{j-1}), X=Tor_1(K_{i-1},V_{j}), \ldots $
This isn't clear from the definition given. What we should have is that $$Tor_1(K_{i-1},V_j) \cong \ker d_1/im \, d_2$$
where with $T=- \otimes_RV_j$, $$ TP_{i+2} \xrightarrow{d_2} TP_{i+1} \xrightarrow{d_1} TP_i \xrightarrow {d_0} K_{i-1} $$
why do the equalities hold? In fact even if it holds, shouldnt it be an isomorphism?
Note that if you have a short exact sequence $0\to K\to P \to L\to 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$\operatorname{Tor}_1(P,M)=0\to \operatorname{Tor}_1(L,M)\to K\otimes M\to P\otimes M\to L\otimes M\to 0$$
so that $\operatorname{Tor}_1(L,M)$ is the kernel of $K\otimes M\to P\otimes M$. This is what Rotman is using throughout.