There is an argument that I would like to fully understand, but I can't see it, yet.
Here is the situation: Given smooth manifolds $X$, $Y$ and $Z$ and transverse maps $f:X\rightarrow Z$, $g\rightarrow Z$, then the fibered product $M:=X\times_{f=g} Y$ is well defined. In fact, it is given by points $(x,y)\in X\times Y$ such that $f(x) = g(y)$. In other words, it is the preiamge of the diagonal $\Delta_Z\subset Z\times Z$ of the map $F=(f,g): X\times Y\rightarrow Z\times Z$.
I am looking for an expression for the tangent bundle $TM$ using the language of $K$-theory (i.e. a virtual bundle). I have read the following argument: The normal bundle of $\Delta_Z$ in $Z\times Z$ is isomorphic to (the pullback of) $TZ$, so it follows that the tangent bundle of the fibered square is $TM = TX+TY - f^*TZ$.
It must have something to do with the transversallity of $f$ and $g$ but I can't fill in the details. Any help is appreciated:)
With some help by Ted Shifrin, I was able to find a solution:
First let $f:X\rightarrow Y$ be smooth, $Z\subset Y$ and $f$ transverse to $Z$. Then it is a well known fact that $W = f^{-1}(Z)$ is a submanifold.
I want to show that $N(W,X) \cong f^*N(Z,Y)$ by defining a suitable bundle isomorphism:
Let $x\in X$ be such that $f(x)\in Z$. It is known see here that $T_xW = (df(x))^{-1}(T_{f(x)}Z)$. Consider the induced map $g: T_x X\rightarrow T_{f(x)}Y/T_{f(x)}Z$, $v\mapsto df(x)[v]+T_{f(x)}Z$. The kernel of this map is $$\text{ker}(g) = (df(x))^{-1}(T_{f(x)}Z) = T_xW.$$ By transversality, we find the image of $g$ to be $$ \text{im}(g) = (df(x)(T_xX) + T_{f(x)}Z)/T_{f(x)}Z = T_{f(x)}Y/T_{f(x)}Z,$$ so we obtain an isomorphism $T_{x}X/T_xW\overset{\sim}{\longrightarrow} T_{f(x)}Y/T_{f(x)}Z$ induced by $df(x)$. This map induces a bundle isomorphism $N(X,W) = TX/TW \cong f^* TY/TZ = f^*N(Z,Y)$.
For the original question: Let $M = F^{-1}(\Delta_Z)$. Denote by $\pi_i:Z\times Z\rightarrow Z$ the projections onto the first and second factor, respectively. The kernel of $d\pi_2$ is exactly $\pi_2^* TZ$ and the restriction of $d\pi_2$ to $T\Delta_Z$ defines an isomprhism to $\pi_2^*TZ$. So $\pi_1^*TZ$ is isomorphic to the normal bundle of $T\Delta_Z$ in $Z\times Z$.
Let $(x,y)$ be such that $f(x) = g(y)$, then transversality gives $$dF(x,y)(T_{(x,y)}X\times Y) + T_{F(x,y)}\Delta_Z = df(x)(T_xX) + dg(y)(T_yY) + T_{F(x,y)}\Delta_Z = T_{f(x)}Z + T_{g(y)}Z = T_{F(x,y)}(Z\times Z),$$ so $F$ is transverse to $\Delta_Z$. Thus, $M$ is indeed a submanifold of $X\times Y$ and $$ T(X\times Y)/TM = N(M,X\times Y) \cong F^*N(\Delta_Z,Z\times Z) \cong F^*\pi_1^* TZ.$$ Since $\pi_1\circ F\circ j_X = f$, where $j_X:X\hookrightarrow X\times Y$, we can write $$ TM = TX + TY - f^*TZ.$$
There might be some minor technical issues but the idea should (hopefully) be correct.