The moment an unknown and unknowable in advance future quantity enters an optimal solution, we have no other option than to insert in its place some estimation of it. The more widely used such estimation (but not the only one) is the conditional expectation. So in reality your (feasible) solution is

$$E\big[f(\alpha_t,\beta_t,s_t,s_{t+1}\mid t\big]=0 \Rightarrow f(\alpha_t,\beta_t,s_t,E[s_{t+1}\mid t]) =0$$

Where the conditional expected value operator progressed directly through everything (non-linearities included), because conditional on $t$ everything else expect $s_{t+1}$ is treated as a constant.

Remember also that the conditional expectation is a function, and not a constant (as is the unconditional expected value).

Since this is a solution, to study comparative statics one has to use the implicit function theorem which states that, starting at the solution,

$$\frac{{\rm d}s_t}{{\rm d}\alpha_t} = -\frac {\partial f(\alpha_t,\beta_t,s_t,E[s_{t+1}\mid t]) / \partial \alpha_t}{\partial f(\alpha_t,\beta_t,s_t,E[s_{t+1}\mid t]) / \partial s_t}$$

The partial derivative symbol in the right-hand side conveys the message that in the numerator, $s_t$ is not to be differentiated with respect to $\alpha_t$, and in the denominator, $\alpha_t$ is not to be differentiated with respect to $s_t$ (and the same holds with respect to $\beta_t$).
But $E[s_{t+1}\mid t]$ is fair game in both cases. What will the operations $\partial E[s_{t+1}\mid t] /\partial \alpha_t$ and $\partial E[s_{t+1}\mid t] /\partial s_t$ yield, will depend of course on what is the actual expression for the conditional expectation.

The moment an unknown and unknowable in advance future quantity enters an optimal solution, we have no other option than to insert in its place some estimation of it. The more widely used such estimation (but not the only one) is the conditional expectation. Conditional on $t$, everything except $s_{t+1}$ will be treated as a constant. Since we do not know how exactly $s_{t+1}$ enters the $f$-function, we write in abstract notation as a function:

$$E\big[f(\alpha_t,\beta_t,s_t,h(s_{t+1},...)\mid t\big]=0 \Rightarrow f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) =0$$

Remember also that the conditional expectation is a function, and not a constant (as is the unconditional expected value).

Since this is a solution, to study comparative statics one has to use the implicit function theorem which states that, starting at the solution,

$$\frac{{\rm d}s_t}{{\rm d}\alpha_t} = -\frac {\partial f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) / \partial \alpha_t}{\partial f(\alpha_t,\beta_t,s_t,E[h(s_{t+1},...)\mid t]) / \partial s_t}$$

The partial derivative symbol in the right-hand side conveys the message that in the numerator, $s_t$ is not to be differentiated with respect to $\alpha_t$, and in the denominator, $\alpha_t$ is not to be differentiated with respect to $s_t$ (and the same holds with respect to $\beta_t$).
But $E[h(s_{t+1},...)\mid t]$ is fair game in both cases. What will the operations $\partial E[h(s_{t+1},...)\mid t] /\partial \alpha_t$ and $\partial E[h(s_{t+1},...)\mid t] /\partial s_t$ yield, will depend of course on what is the actual expression for the conditional expectation.