I guess that in $W_{\alpha \beta}$ the indices $i$ and $j$ are here but not written and are contracted in $W^2$ in the same way that in the first $W^{2}$. So I can forget these indices in what follows.

I will show that the two $W^{2}$ are the same if we have the relation

$W_{\mu\nu} =\frac{1}{2}W_{\alpha\beta}(\sigma_\mu)^{\alpha\dot\alpha}(\sigma_\nu)^{\beta\dot\beta}\varepsilon_{\dot\alpha\dot\beta}$

which is the same thing suggested by the first equation of the question up to the factor 1/2. I don't know if this factor is relevant. Given this relation, we have

$W_{\mu\nu}W^{\mu\nu} = \frac{1}{4}W_{\alpha\beta}W_{\alpha'\beta'}(\sigma_\mu)^{\alpha\dot\alpha}(\sigma^\mu)^{\alpha'\dot\alpha'} (\sigma_\nu)^{\beta\dot\beta} (\sigma^\nu)^{\beta'\dot\beta'} \varepsilon_{\dot\alpha\dot\beta}\varepsilon_{\dot\alpha'\dot\beta'}$.

Now the key point is the identity (*) $(\sigma_\mu)^{\alpha\dot\alpha}(\sigma^\mu)^{\alpha'\dot\alpha'} = 2 \varepsilon^{\alpha \alpha'}\varepsilon_{\dot\alpha \dot\alpha'}$. Using (*) and the analoguous relation with the $\alpha$'s replaced by the $\beta$'s, and simplifying the $\epsilon$'s (we have $\varepsilon^{\dot\alpha\dot\alpha'}\varepsilon_{\dot\alpha\dot\beta}\varepsilon^{\dot\beta\dot\beta'} \varepsilon_{\dot\alpha'\dot\beta'}=1$), we obtain

$W_{\mu\nu}W^{\mu\nu}= W_{\alpha\beta}W_{\alpha'\beta'} \varepsilon^{\alpha\alpha'}\varepsilon^{\beta\beta'}$.

Proof of (*): the identity (*) is a part of a large class of identities generally referred to as Fierz identities. These identities are all consequences of the fact that the four matrices $\sigma^\mu$ are a basis of the four dimensional vector space of two by two complex matrices. See for example this question:

http://www.physicsoverflow.org/17763/how-does-one-prove-fierz-identities?show=17763#q17763

Using $\sigma^0 = 1$, $Tr(\sigma^i)=0$, and $Tr(\sigma^i \sigma^j)=2 \delta^{ij}$ for $i,j=1,2,3$, we find that any two by two complex matrix $M$ can be written $M = \frac{1}{2} Tr(M) 1 + \frac{1}{2}Tr(M \sigma^i) \sigma^i$. Applying this to $M$ being the matrix $M_{\alpha \dot\alpha}(\beta \dot\beta)=\delta_{\alpha \beta} \delta_{\dot \alpha \dot\beta}$ (the matrices $M(\beta \dot\beta)$ form a natural basis of the space of two by two complex matrice) gives

$\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta}= \frac{1}{2} \delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha} + \frac{1}{2}(\sigma^i)_{\dot\beta \beta}(\sigma^i)_{\alpha \dot\alpha}$. Therefore,

$-(\sigma^i)_{\dot\beta \beta}(\sigma^i)_{\alpha \dot\alpha}=\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha} -2\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta} $.

In signature (+---), we have $(\sigma^\mu)_{\alpha \dot\alpha} (\sigma^\mu)_{\dot\beta \beta}=\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha} -(\sigma^i)_{\beta \dot\beta}(\sigma^i)_{\alpha \dot\alpha}$. Combining with the previous equality gives

$(\sigma^\mu)_{\alpha \dot\alpha} (\sigma^\mu)_{ \dot\beta \beta}= 2 (\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha}-\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta})=2 \varepsilon_{\alpha \dot\beta}\varepsilon_{\dot\alpha \beta}$, hence (*).