For lm() coefficient in R, why not give slope directly? Focus "slope",not a parameter estimation method Like this example, it shows (Intercept) and x, why not slope?  And I'm not sure x means what?
x <- c(1,4,12,34,86,99)

y <- c(12, 48, 500, 1000, 1200,3242)  

plot(x, y)  

lm(y~x)
Coefficients:
(Intercept)            x  
      27.65        24.73  

summary(lm(formula = y ~ x))

Call:
lm(formula = y ~ x)

Residuals:
      1       2       3       4       5       6 
-40.38  -78.57  175.60  131.56 -954.36  766.16 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)   27.652    360.332   0.077   0.9425  
x             24.729      6.487   3.812   0.0189 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 623.2 on 4 degrees of freedom
Multiple R-squared:  0.7842,    Adjusted R-squared:  0.7302 
F-statistic: 14.53 on 1 and 4 DF,  p-value: 0.0189

 A: The issue is, that lm() or summary() do have different labelling conventions for each explanatory variable.
Instead of writing slope_1, slope_2,.... They simply use the name of the variable in any output to indicate which coefficients belong to which variable.
Extending your example for another variable
x1 <- c(1,4,12,34,86,99)
x2 <- c(10,11,0,4,8,9)

y <- c(12, 48, 500, 1000, 1200,3242) 

lm(y~x1+x2)
summary(lm(y~x1+x2))

...
Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  138.968    647.456   0.215   0.8438  
x1            24.946      7.492   3.330   0.0447 *
x2           -17.122     76.737  -0.223   0.8378  
...

Outputs the two slopes of the regression - x1 and x2 - and the intercept.
A: Here is a title--- a simple example of a first degree spline with single knot and interpretation of the estimated coefficients to calculate the slope of the fitted lines in https://stackoverflow.com/questions/37362738/how-to-interpret-lm-coefficient-estimates-when-using-bs-function-for-splines
I think this regression uses two E(y)=α+xβ, so, needn't calculate these two slopes (when degree = 1), according to what I have learnt, just extract from summary is okay, but I'm not sure...
fit <- lm(formula = y ~ bs(x, degree = 1, knots = c(0)))

Coefficients: 
                                 Estimate Std. Error t value Pr(>|t|)   
   
(Intercept)                       5.01351    0.02072   242.0   <2e-16 *** 
bs(x, degree = 1, knots = c(0))1 -5.04119    0.03034  -166.2   <2e-16 ***
bs(x, degree = 1, knots = c(0))2  4.96375    0.02714   182.9   <2e-16 ***

